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HIGHER GHOMETRY 


TP RgieG ON: OOM Te ie: 


THIRD PART OF A SERIES 
ON 


ELEMENTARY AND HIGHER 


GEOMETRY, TRIGONOMETRY, AND MENSURATION, 


CONTAINING MANY VALUABLE DISCOVERIES AND IMPROVEMENTS IN MATHEMATICAL 
SCIENCE, ESPECIALLY IN RELATION TO THE QUADRATURE OF THE CIRCLE, 
AND SOME OTHER CURVES, AS WELL AS THE CUBATURE OF CERTAIN 
CURVILINEAR SOLIDS 5; DESIGNED AS A TEXT-BOOK FOR COLLEGIATE 
AND ACADEMIC INSTRUCTION, AND AS A PRACTICAL 
COMPENDIUM OF MENSURATION. 


BY NATHAN *SGHOLFIELD. 


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NEW YORK: 


PUBLISHED BY COLLINS, BROTHER & CO. 
No..254 Pearl Street. 


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1845, 


Entered according to Act of Congress, in the year 1845, by 


NATHAN SCHOLFIELD, 


In the Clerk’s Office of the District Court of Connecticut. 
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G. W. WOOD, PRINTER, 29 GOLD ST., NEW YORK. 
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Sch] \. aTHEMaTics 
LIBRARY, 


PREFACE. 


Tuts part of the series consists of spherical geometry, takgn 
mostly from Brewster’s translation of Legendre’s work. Ana- 
lytical plane and spherical trigonometry, based on the subject, 
as found in Rutherford’s edition of Hutton’s Mathematics, being 
originally abridged from the larger works of Cagnoli, and 
others ; but, in this work, much improved and enlarged. To 
which are added many practical exercises on the subject, by 
way of application. In this treatise will be found many cu- 
rious and highly useful problems in trigonometrical surveying, 
and topographical operations, not before published. The pro- 
perties of the circle are introduced advantageously into trigo- 
nometrical problems—hence we are enabled, by geometrical 
construction, and trigonometrical analysis, to determine many 
otherwise extremely difficult problems, in a manner at once 
simple, elegant, and satisfactory. The application of alge- 
bra to geometry, is discussed in such manner as to combine 
the principles of the two sciences. The properties of the pa- 
rabolic, elliptical, and hyperbolic curves, being such as are 
formed by the sections of a cone, and hence are usually de. 
nominated conic sections, are also discussed. This subject is, 
with some alterations and additions, taken from Rutherford’s 
edition of Hutton’s Mathematics. It is the design of the au- 
thor, to preserve an unbroken connection from pure elemen- 
tary to the higher Geometry and mensuration; and with this 
object in view the present volume, being the third part of the 
series, is prepared. 


Eee Bis. 


iV PREFACE. 


The well established reputation and the high respectability 
of the authors from whom our selections have been made, 
renders it unnecessary for us to discuss their merits in order 
to secure a favorable reception of this. It will only be neces- 
sary for us, in the following pages, to preserve the same de- 
gree of accuracy and perspicuity in our digressions as charac- 
terise those works, and we shall have nothing to fear from the 
criticisms of scientific amateurs and mathematicians. 

’ 


CONTENTS. 


SPHERICAL GEOMETRY. 


| PAGE. 
Definitions and General Propositions, - - - - -— = 9 
ANALYTICAL PLANE TRIGONOMETRY. 
Cuap. I.—Definitions and Illustration of Principles, - - - 27 
Cuap. II.—General Formule, - . . - > - - 45 
Cuap. III.—Formule for the Solution of Triangles, - - - 66 
Cuar. [V.—Construction of Trigonometrical Tables, - - - 738 
Cuap. V.—Logarithms, aay he Ot = SS Lickers helt Heike =). See ae 
Cuar. VI.—Solution of Right Angled Triangles, aS Ate grag eae OS 
Cuap. VII.—Solution of Oblique Angled Triangles, - - - 98 
Cuap. VIII.—On the use of Subsidiary Angles, - - - = 107 
Cuap. [X.—On the Solution of Geometrical Problems by Trigo- 
nometry, - - - - - - . oo DU 
Cuap. X.—Problems in Trigonometrical Surveying, &c. - - 113 
SPHERICAL TRIGONOMETRY. 
Cuar. I.—General Principles and Illustrations, - - - - 131 
Cuap. II.—Solution of Right Angled Spherical Triangles,- - 144 
Culp. Ill.—Solution of Oblique Angled Spherical Triangles, - 150 


Cuap. IV.—On the use of Subsidiary Angles, - - - = 154 


vi . CONTENTS. 


APPLICATION OF ALGEBRA TO GEOMETRY. 


PAGE. 

Construction of Algebraical Quantities, - - - - - 158 
Geometrical Questions, the modes of forming Equations therefrom, 

and their Solutions, - - - - + - - 166 


Determination of Algebraic Expressions for Surfaces and Solids, 180 


CONIC SECTIONS. 
The Parabola and its Properties, oer, ay mad Bias - - 189 
The Ellipse and its Properties, - - - - - sue Ars 


The Hyperbola and its Properties, -  - a Ai Fe at - 215 


( 


SPHERICAL GEOMETRYS: ©. 


SPHERICAL GEOMETRY. 
DEFINITIONS. 


1. Tue sphere is a solid terminated by a curve surface, all 
the points of which are equally distant from a point within, 
called the centre. 


The sphere may be con- 
ceived to be generated by 
the revolution of a semi- 
circle DAE about its di- 
ameter DE; forthe surface 
described in this move- 
ment, by the curve DAE, 
will have all its points 
equally distant from its 
centre C. 


2. The radius of a sphere is a straight line, drawn from the 
centre to any point of the surface ; the diameter, or axis, is a 
line passing through this centre, and terminated on both sides 
by the surface. 

All the radii of a sphere are equal; all the diameters are 
equal, and each double of the radius. 

3. It will be shown (Prop. I.) that every section of the 
sphere, made by a plane, is a circle. This granted, a great 
circle is a section which passes through the centre; a small 
circle, one which does not pass through the centre. 

4. A plane is tangent to a sphere, when their surfaces have 
but one point in common. 


a { . 


hi. 


8 SPHERICAL GEOMETRY. 


5. The pole of a circle of a sphere is a point’ in the surface 
equally distant from all the points in the circumference of this 
circle. ; 

6. A spherical triangle isa portion of the surface of asphere, 
bounded by three arcs of great circles. 

Those arcs, named the sides of the triangle, are always 
supposed to be each less than asemi-circumference. The an- 
gles which their planes form with each other, are the angles 
of the triangle. 

7. A spherical triangle takes the name of right-angled, 
isosceles, equilateral, in the same cases as a rectilineal triangle. 

8. A spherical polygon is a portion of the surface of a 
sphere, terminated by several arcs of great circles. 

9. A lune is that portion of the surface of a sphere which 
is included between two great semicircles, meeting in a com- 
mon diameter. ; 

10. A spherical wedge, or ungula, is that portion of the solid 
sphere which is included between the same great semi-circles, 
and has the lune for its base. 

11. A spherical pyramid is a portion of the solid sphere in- 
cluded between the planes of a solid angle, whose vertex is 
the centre. The base of the pyramid is the spherical polygon 
intercepted by the same planes. 

12. A zone is the portion of the surface of the sphere in- 
cluded between two parallel planes, which form its bases. One 
of those planes may be tangent to the sphere ; in which case, 
the zone has only a single base. 

13. A spherical segment is the portion of the solid sphere 
included between two parallel planes which form its bases. 

One of these planes may be tangent to the sphere ; in which 
case, the segment has only a single base. 

14. The altitude of a zone or of a segment is the distance 
between the two parallel planes, which form the bases of the 
zone or segment. 

15. Whilst the semicircle DAE (Def. 1.) revolving round 
its diameter DE, describes the sphere, any circular sector, as 
DCF or FCH, describes a solid, which is named a spherical 
sector. 

16. The symbal °.* which occurs in this volume, is used to 
denote because ; when applied in algebraic notation. 


. 
. ie 


SPHERICAL GEOMETRY. 9 
PROPOSITION I. THEOREM. 
Every section of a sphere, made by a plane, is a circle. 
q © 


Let AMB be a section, made by a D 
plane, in the sphere, whose centre is C. YEE 
From the point C, draw CO perpendicu- 4 — B 
lar to the plane AMB ; and different lines 
CM, CM, to different points of the curve 
AMB, ohich terminates the section. 

The oblique lines CM, CM, CA, are 
equal, being radii of the sphere; hence 
(Prop. VI. B. I. El. S. Geom.) they are 
equally distant from the perpendicular CO; therefore all the 


lines OM, MO, OB, are equal. Consequently, the section 
AMB isa circle, whose centre is O. 


Cor. 1. If the section passes through the centre of the 
sphere, its radius will be the radius of the agile ; hence, all 
great circles are equal. 


Cor. 2. ‘Two great circles always bisect each other ; for 
their common intersection, passing through the centre, is a 
diameter. 


Cor. 3. Every great circle divides the sphere and its sur- 
face into two equal parts; for, if the two hemispheres were 
separated, and afterwards placed on the common base, with 
their convexities turned the same way, the two surfaces would 
exactly coincide, no point of the one being nearer the centre 


than any point of the other. 


Cor. 4. The centre of a small circle, and that of the 
sphere, are in the same straight line, perpendicular to the plane 
of the small circle. 


Cor. 5. Small circles are the less the further they lie from 
the centre of the sphere ; for, the greater CO is, the less is the 
chord AB, the diameter of the small circle AMB. 


Cor. 6. Anarc of a great circle may always be made to 
pass through any two given points of the surface of the 
sphere; for the two given points, and the centre of the sphere, 
make three points, which determine the position of a plane. 
But if the two given points were at the extremities of a di- 
ameter, these two points and the centre would then lie in one 
straight line, and an infinite number of great circles might be 
made to pass through the two given points. 


10 SPHERICAL GEOMETRY. 
PROPOSITION Il. THEOREM. 


In every spherical triangle, any side is less than the sum of the 
' + other two. 


Let O be the centre of the sphere, y 
and ACB the triangle: draw the 
radii OA, OB, OC. Imagine the 
planes AOB, AOC, COB, to be | 
drawn; those planes will form a 
solid angle at the centre O; and the 
angles AOB, AOC, COB, wiil be 
measured by AB, AC, BC, the sides 
of the spherical triangle. But each of 
the three plane angles forming a solid 
angle is less than the sum of the other two (Prop. XXI., B. I. 
El. S. Geom.): hence any side of the triangle ABC is less 
than the sum of the other two. 


oy 


we] 


PROPOSITION Ill, THEOREM. 


The shortest distance from one point to another, on the surface 
of a sphere, is the arc of the great circle which joins the two 
given points. 


“ Let ADB be the arc of the great circle 
which joins the points A and B; and with- 
out this line, if possible, let M be a point of 
the shortest path between A and B. Through 
the point M, draw MA, MB, arcs of great ay D 
circles; and take BD=MB. 

By the last Proposition, the arc ADB is 
shorter than AM+MB; take BD=BM re- 
spectively from both; there will remain B 
AD<AM. Now, the distance of B from- 
M, whether it be the same with the are BM, or with any 
other line, is equal to the distance of B from D; for, by ma- 
king the plane of the great circle BM to revolve about the 
diameter which passes through B, the point M may be brought 
into the position of the point D); and the shortest line between 
M and B, whatever it may be, will then be identical with that 
between D and B: hence the two paths from A to B, one 
passing through M, the other through D, have an equal part in 
each, the part from M to B equal to the part from D to B. 
The first part is the shorter by hypothesis; hence the distance 


~ Pe id . “ee bg P ad 
mae SPHERICAL GEOMETRY. — 11 


from A to M must be shorter than the distance’ from A to D, 
which is absurd; the arc AM being proved greater than AD. 
Hence no point of the shortest line from A to B can lie out of 
the arc ADB; hence this arc is itself the shortest distance be- 
tween its two extremities. 


PROPOSITION IV. THEOREM, 


The sum of the three sides of a spherical triangle is less than 
the circumference of a great circle. 


Let ABC be any spherical tri- C 
angle; produce the sides AB, 
AC, till they meet again in D. 
The arcs ABD, ACD, will be 
semicircumferences, since (Prop. D 


I. Cor. 2.) two great circles al- 
ways bisect each other. But in 

the triangle BCD, we have (Prop. ,’ 
II.) the side BC<BD+CD; add 
AB+AC to both; we shall have 
AB+AC+BC<ABD+ACD :— 
that is to say, less than a circum- 
ference. 


PROPOSITION V. THEOREM. 


The sum of all the sides of any spherical polygon is less than 
the circumference of a great circle. 


Take the pentagon ABCDE, for C 
example. Produce the sides AB,’ D 
DC, till they meet in F ; then, since 
BC is less than BF+CF, the peri- © B 
meter of the pentagon ABCDE will 
be less than that of the quadrilateral E 
AEDF. Again produce the sides A. 
AE, FD, till they meet in G; we 
shall have ED<EG+DG;; hence the perimeter of the quadri- 
lateral AEDF is less than that of the triangle AFG; which 
last is itself less than the circumference of a great circle: 
hence, for a still stronger reason, the perimeter of the polygon 
ABCDE is less than this same circumference. 


ry 


Scholium. This proposition is fundamentally the same as 


Prop. XXII. B. I. Ll. S. Geom. ; for O being the centre of the 


~ 


12 SPHERICAL GEOMETRY. 


sphere, a solid angle may be conceived as formed at O, by 
the plane angles AOB, BOC, COD, &c. ; and the sum of these 
angles must be Jess than four right angles; which is exactly 
the proposition we have been engaged with. The demon- 
stration here given is different from that of Prop. XXII. B. 
I. El. S. Geom.; both, however, suppose that the polygon 
ABCDE is convex, or that no side produced will cut the 
figure. 


PROPOSITION VI. THEOREM. 


The poles of a great circle of the sphere, are the extremities of 
that diameter of the sphere which is perpendicular to this cir- 
cle ; and these extremities are also the poles of all small circles 
parallel to it. 


Let ED be perpendi- 
cular to the great circle 
AMB;; then will E and 
D be its poles; as also 
the poles of the parallel 
small circles HPP, FNG. 

For, DC being perpen- 
dicular to the plane AMB, 
is perpendicular to all the 
straight lines CA, CM, CB, 
&c., drawn through its 
foot in this plane; hence 
all the arcs DA, DM, DB, 
é&c., are quarters of the 
circumference. So_like- . E 
wise are all the arcs EA, EM, EB, &c.; hence the points D 
and i; are each equally distant from all the points of the cir- 
cumference AMB; hence (Def. 5.) they are the poles of that 
circumference. 

Again, the radius DC, perpendicular to the plane AMB, is 
perpendicular to its parallel ENG; hence (Prop. I. Cor. 4.) it 
passes through O the centre of the circle FNG; hence, if the 
oblique lines DF, DN, DG be drawn, these oblique lines will 
diverge equally’ from the perpendicular DO, and will them- 
selves be equal. But, the chords being equal, the arcs are 
equal ; hence the point D is the pole of the small circle FNG; 
and, for like reasons, the point E is the other pole. 


Cor.1. HKivery arc DM, drawn from a point in the arc of 
a great circle AMB to its pole, is a quarter of the circumfer- 


SPHERICAL GEOMETRY. 13 


ence, which, for the sake of brevity, is usually named a quad- 
rans, or quadrant: and this quadrant at the same time makes 
aright angle with the arc AM. For (Prop. XVII. B. I. Ei. 
S. Geom.) the line DC being perpendicular to the plane AMC, 
every plane DMC passing through the line DC, is perpendicu- 
lar to the plane AMC; hence the angle of these planes, or 
(Def. 6.) the angle AMD, is a right angle. 


Cor. 2. To find the pole of a given arc AM, draw the in- 
definite are MD perpendicular to AM; take MD equal to a 
quadrant; the point D will be one of the poles of the are AM: 
or thus, at the two points A and M, draw the arcs AD and MD 
perpendicular to AM; their point of intersection, D, will be 
the pole required. 


Cor. 3. Conversely, if the distance of the point D from 
each of the points A and M, is equal to a quadrant, the point 
D will be the pole of the arc AM, and also the angles DAM, 
AMD, will be right. 

For, let C be the centre of the sphere, and draw the radii 
CA,CD,CM. Since the angles ACD, MCD are right, the 
line CD is perpendicular to the two straight lines CA, CM ; 
hence it is perpendicular to their plane (Prop. V. B. I. Ei. S. 
Geom.) ; hence the point D is the pole of the are AM; and 
consequently the angles DAM, AMD are right. 


Scholium. The properties of these poles enable us to de- 
scribe arcs of a circle on the surface of a sphere, with the 
same facility as ona plane surface. It is evident, for instance, 
that by turning the arc DF’, or any other line extending to the 
same distance, round the point D, the extremity F will de- 
scribe the small circle ENG; and, by turning the quadrant 
DFA round the point D, its extremity A will describe the are 
of the great circle AM. 

If the arc AM were required to be produced, and nothing 
were given but the points A and M through which it was to 
pass, we should first have to determine the pole D, by the in- 
tersection of two ares described from the points A and M as 
centres, with a distance equal toa quadrant. The pole D 
being found, we might describe the arc AM and its prolonga- 
tion, from D as a centre, and with the same distance as before. 

In fine, if it is required from a given point P to let fall a 
perpendicular on the given are AM, produce this arc to 8, 
till the distance PS be equal to a quadrant; then, from the 
pole S, and with the same distance, describe the are PM, which 
will be the perpendicular required. 

9* 


14 SPHERICAL GEOMETRY. 
PROPOSITION VII. comtiee si 


Every plane perpendicular to a radius at its extremity is tan- 
gent to the sphere. 


Let FAG (sce the next diagram) be a plane perpendicular to 


the radius OA, at its extremity A. Any point M in this plane 


being assumed, and OM, AM being joined, the angle OAM 
will be right, and hence the distance OM will be greater than 
OA. Hence the point M lies without the sphere; and as the 
same can be shown for every other point of the plane FAG, 
this plane can have no point but A common to it and the sur- 
face of the sphere ; hence (Def. 4.) it is tangent. 


Scholium. In the same way, it may be shown that two 
spheres have but one point in common, and therefore touch 
each other, when the distance between their centres is equal to 
the sum, or the difference of their radii; in which case, the 
centres and the point of contact lie in the same straight line. 


PROPOSITION VIlI THEOREM. 


The angle formed by two arcs of great circles, is equal to the 
angle formed by the tangents of these two arcs at their points 
of intersection, and ts measured by the arc described from 
this point of intersection, as a pole, and limited by the sides, 
produced if necessary. 


Let the angle BAC be formed by the two 
arcs AB, AC; then it will be equal to the 
angle FAG formed by the tangents AF’, AG, 
and be measured by the arc DE, described 
about A as a pole. 

For the tangent AF’, drawn in the plane 
of the arc AB, is perpendicular to the ra- 
dius AO; and the tangent AG, drawn in 
the plane of the arc AC, is perpendicular to 
the same radius AO. Hence, (Prop. XX. 
B. J. £7. S. Geom.) the angle AG is equal 
to the angle contaimed by the planes OAB, 
OAC; which is that of the ares AB, AC, Hi 
and is named BAC. 

In like manner, if the arcs AD and AE are both quadrants, 
the line OD, OF will be perpendicular to OA, and the angle 
DOE will still be equal to the angle of the planes AOD, AOE: 
hence the arc DE is the measure of the angle contained by 
these planes, or of the angle CAB. 


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_. SPHERICAL GEOMETRY. 15 


Cor. The angles of spherical triangles may be compared 
together, by means of the arcs of great circles described from 
their vertices as poles and included between their sides : hence 


itis easy to make an angle of this kind equal to a given angle. 


. BC, AG, AB. 


Scholium.. Vertical angles, such as ACO and BCN (see dia- 
gram to Prop. X XI.) are equal; for either of them is still the 
angle formed by the two planes ACB, OCN. 

It is further evident, that, in the intersection of two arcs 
ACB, OCN, the two adjacent angles ACO, OCB taken toge- 
ther, are equal to two right angles. 


PROPOSITION IX. THEOREM. 


If from the vertices of the three angles of a spherical triangle, 
as poles, three arcs be described forming a secund triangle, 
the vertices of the angles of this second triangle will be re- 


spectively poles of the sides of the first. 


From the vertices A, B, C, D 
as poles, let the arcs EF, FD, 
ED be described, forming on 
the surface of the sphere, the 
triangle DFE; then will the 
points D, E, and F be re- E 
spectively poles of the sides 


For, the point A being the 
pole of the arc EF’, the dis- 
tance AF is a quadrant; the 
point C being the pole of the arc DE, the distance CE is like- 


Wise a quadrant: hence the point EF. is removed the length of # 


a quadrant from each of the points A and C; hence (Prop VI. 
Cor. 3.) is the pole of the arc AC. It might be shown, by the 
same method, that D is the pole of the arc BC, and F that of 
the arc AB. 


Cor. Hence the triangle ABC may be described by means of 
DEF, as DEF is described by means of ABC. > 


PROPOSITION X. THEOREM. 

The same supposition continuing as in the last proposition, each 
angle in the one of the triangles, will be measured by the 
semicircumference minus the side lying opposite to wt in the 


other triangle. 


Produce the sides (see the preceding diagram) AB, AC, if 


16 “#4 SPHERICAL GEOMETRY. 


~ necessary, till they meet EF, inG and H. The point A being 


the pole of the arc GH, the angle A will be measured by that 


arc. But the arc EH is a quadrant, and likewise GI, E be- 
ing the pole of AH, and F of AG; hence EH+GF is equal 
to the semicircumference. Now, EH+GEF is the same as 
EF+GH; hence the arc GH, which measures the angle A, 
is equal to a semicircumference minus the side KF’. In like 
manner, the angle B will be measured by 3 ctre.—DF: the 
angle C, by 1 circ.—DE. 

And this property must be reciprocal in the two triangles, 
since each of them is described in a similar manner by means 
of the other. Thus we shall find the angles D, E, F, of the 
triangle DEF to be measured respectively by 4 circ.—BC, 
1 circ.—AC. 1 circ—AB. Thus the angle D, for example, 
is measured by the arc MI; but MI+BC=MC+BI=1} circ. ; 
hence the arc MI, the measure of D, is equal to 4 circ.—BC ; 
and so ofall the rest. 


Scholium. It must further be observed, that besides the tri- 
angle DEF’, three others might be formed by the intersection 
of the three arcs DE, EF, DF. But the 
proposition immediately before us is ap- D 
plicable only to the central triangle, 
which is distinguished from the other ) 
three by the circumstance (see diagram / 
to Prop. IX.) that the two angles A and 
D lie on the same side of BC, the two B 
and FE on the same side of AC, and the d 
two C and F on the same side of AB. 

Various names have been given to the triangles ABC, DEF ; 
we shall call them polar triangles. 


PROPOSITION XI. THEOREM. 


If around the vertices of the two angles of a given spherical trt- 
angle, as poles, the crcumference of two circles be described 
which shall pass through the third angle of the triangle ; if 
then, through the other point in which those circumferences 
intersect, and the two first angles of the triangle, the arcs of 
great circles be drawn, the triangle thus formed will have all 
its parts equal to those of the first triangle. 


Let ABC be the given triangle, CED, DFC the arc de- 
scribed about A and B as poles; then will the triangle ADB 
have all its parts equal to those of ABC. 


SPHERICAL GEOMETRY. 17 


For, by construction, the side AD=AC, A 
DB=BC, and AB is common; hence 
those two triangles have their sides equal, /\ 
each to each. We are now to show, that 


O 
the angles opposite these equal sides are F 
also equal. ¥ 
If the centre of the sphere is supposed © Re 
to be at O, a solid angle may be conceiv- E 
ed as formed at O by the three plane an- 
gles AOB, AOC, BOC ; likewise another 5 


solid angle may be conceived as formed by the three plane 
angles AOB, AOD, BOD. And because the sides of the tri- 
ongle ABC are equal to those of the triangle ADB, the plane 
angles forming the one of these solid angles, must be equal to 
the plane angles forming the other, each to each. But in this 
case we have shown (Prop. XXIII. Hl. S. Geom.) that the 
planes, in which the equal angles lie, are equally inclined to 
each other ; hence, all the angles of the spherical triangle DAB 
are respectively equal to those of the triangle CAB, namely, 
DAB=BAC, DBA=ABC, and ADB—=ACB;; hence, the sides 
and the angles of the triangle ADB are equal to the sides and 
the angles of the triangle ACB. 


Scholium. The equality of those triangles is not, however, 
~ an absolute equality, or one of superposition ; for it would be 
impossible to apply them to each other exactly, unless they 
were isosceles. The equality meant here is what we have al- 
ready named an equality by symmetry ; therefore, we shall call 
the triangles ACB, ADB, symmetrical triangles. 


PROPOSITION XII. THEOREM. 


Two triangles on the same sphere, or on equal spheres, are equal 
in all their parts, when they have each an equal angle included 
between equal sides. 


Suppose the side AB=EF, the ie 
side AC=EG, and the angle BAC E 
=I EG; the triangle EG may be 
placed on the triangle ABC, or on 
ABD symmetrical with ABC, just 
as two rectilineal triangles are 
placed upon each other, when they p 
have an equal angle included be- C 
tween equal sides. Hence all the . 
parts of the triangle EF'G will be B F 
equal to all the parts of the trian- 


2 ’ “F 


18 SPHERICAL GEOMETRY. 


gle ABC ; that is, besides the three parts equal by hypothesis, 
we shall have the side BC=FG, the angle ABC=EFG, and 
the angle ACB=EGF. inks 


“ 


PROPOSITION XIII. THEOREM. 


Two triangles on the same sphere, or on equal spheres, are equal 
an all their parts, when two angles and the included side of the 
one are respectively equal to two angles and the included side 
of the other. 


For, one of those triangles, or the triangle symmetrical with 
it, may be placed on the other, as is done in the corresponding 
case of rectilineal triangles, (Prop. IX. B. Il. El. Geom.) 


PROPOSITION XIV. THEOREM. 


If two triangles on the same sphere, or on equal spheres have all 
their sides respectively equal, their angles will likewise be all 
respectively equal, the equal angles lying opposite the equal 
sides. : 


This truth is evident from Proposition XI, where it is shown 
that, with three given sides AB, AC, BC, (see the diagram,) 
there can only be two triangles ACB, ABD, differing as to the 
position of their parts, and equal as to the magnitude of those 
parts. Hence, those two triangles, having all their sides re- 
spectively equal in both, must either be absolutely equal, or at 
least symmetrically so; in both of which cases their corres- 
ponding angles must be equal, and lie opposite to equal sides. 


PROPOSITION XV. THEOREM. 


In every isosceles spherical triangle, the angles opposite the equal 
sides are equal; and conversely, if two angles of a spherical 
triangle are equal, the triangle is isosceles. 


First. Suppose the side AB=AC; we shall 
have the angle C=B. For, if the arc be 
drawn from the vertex A to the middle point 
D of the base, the two triangles ABD, ACD 
will have all the sides of the one respectively 
equal to the corresponding sides of the other, 
namely, AD common, BD=DC, and AB= 
AC: hence, by the last Proposition, their an- 
gles will be equal; therefore B=C. 

Secondly. Suppose the angle B=C; we 


a a 


SPHERICAL GEOMETRY. 19 


shall have the side AC=AB. For, if not, let AB be the greater 
of the two; take BO=AC, and join OC. The two sides BO, 
BC are equal to the two AC, BC; the angle OBC, contained 
by the first two is equal to ACB contained by the second two. 
Hence (Prop. XII.) the two triangles BOC, ACB have all their 
other parts equal; hence the angle OCB=ABC: but by hy- 
pothesis, the angle ABC=ACB; hence we have OCB=ACB, 
which is absurd; hence it is absurd to suppose AB different 
from AC ; hence the sides AB, AC, opposite to the equal an- 
gles B and C, are equal. 


Scholium. The same demonstration proves the angle BAD 
=DAC, and the angle BDA=ADC. Hence the two last are 
right angles; hence the arc drawn from the vertex of an isos- 
celes spherical triangle to the middle of the base, is at right an- 
gles to the base, and bisects the vertical angle. 


PROPOSITION XVI. THEOREM. 


In any spherical triangle, the greater side is opposite the great- 
erangle; and conversely, the greater angle is opposite the 
greater side. 


Let the angle A be greater than the angle B, then will BC 
be greater than AC; and conversely, if BC is greater than 
AC, then will the angle A be greater than B. 


First. Suppose the angle A 
A>B; make the angle BAD= 
B: then (Prop. XV.) we shall 
have AD=DB; but AD+DC 
is greater than AC; hence, 
putting DB in place of AD, we 
shallhave DB+ DC, or BC> AC. 

Secondly. If we suppose BC>AC, the angle BAC will be 
greater than ABC. For, if BAC were equal to ABC, we should 
have BC=AC; if BAC were less than ABC, we should then, 
as has just been shown, find BC<AC. Both these conclusions 
are false: hence the angle BAC is greater than ABC. 


B Cc 


20 SPHERICAL GEOMETRY. 


PROPOSITION XVII. THEOREM. 


If two triangles on the same sphere, or on equal spheres, are 
mutually equiangular, they will also be mutually equilate- 
ral. 


Let A and B be the two given triangles; P and Q their 
polar triangles. Since the angles are equal in the triangles 
A and B, the sides will be equal in their polar triangles P and 
Q, (Prop. X.:) but since the triangles P and Q are mutually 
equilateral, they must also (Prop. XIV.) be mutually equian- 
gular ; and, lastly, the angles being equal in the triangles P 
and Q, it follows (Prop. X.) that the sides are equal in their 
polar triangles A and B. Hence the mutually equiangular 
triangles A and B are at the same time mutually equilateral. 


Scholium. This proposition is not applicable to rectilineal 
triangles ; in which equality among the angles indicates only 
proportionality among the sides. Nor is it difficult to account 
for the difference observable, in this respect, between spheri- 
cal and rectilineal triangles. In the proposition now before 
us, as well as in Propositions XII, XIII, XIV, which treat of 
the comparison of triangles, it is expressly required that the 
arcs be traced on the same sphere, or on equal spheres. Now 
similar arcs are to each other as their radii ; hence, on equal 
spheres, two triangles cannot be similar without being equal. 
Therefore it is not strange that equality among the angles 
should produce equality among the sides. 

The case would be different, if the triangles were drawn 
upon unequal spheres ; there, the angles being equal, the tri- 
angles would be similar, and the homologous sides would be 
to each other as the radii of their spheres. 


PROPOSITION XVIII. THEOREM. 


The sum of all the angles in any spherical triangle is less than 
six right angles, and greater than two. 


For, in the first place, every angle of a spherical triangle 
is less than two right angles (see the following Scholium) : 
hence the sum of all the three is less than six right angles. 

Secondly, the measure of each angle of a spherical triangle 
(Prop. X.) is equal to the semicircumference minus the corres- 
ponding side of the polar triangle; hence the sum of all the 
three, is measured by three semicircumferences minus the sum 


of all the sides of the polar triangle. Now (Prop. IV.), this 


SPHERICAL GEOMETRY. 21 


latter sum is less than a circumference ; therefore, taking it 
away from three semicircumferences, the remainder will be 
greater than one semicircumference, which is the measure of 
two right angles ; hence, in the second place, the sum of all 
the angles in a spherical triangle is greater than two right 
angles. 


Cor. 1. The sum of all the angles of a spherical triangle 
is not constant, like that of all the angles of a rectilineal tri- 
angle; it varies between two right angles and six, without 
ever arriving at either of these limits. Two given angles 
therefore do not serve to determine the third. 


Cor. 2. A spherical triangle may have two, or even three 
angles, right, two or three obtuse. 


If the triangle ABC is bi-rectangular, in other A 
words, has two right angles B and C, the vertex 
A will (Prop. X.) be the pole of the base BC; 
and the sides AB, AC will be quadrants. 

If the angle A is also right, the triangle ABC 
will be tri-rectangular; its angles will all be right, p Cc 
and its sides quadrants. The tri-rectangular triangle is con- 
tained eight times in the surface of the sphere. 


Scholium. In all the preceding observations, we have sup- 
posed, in conformity with (Def. 6,) that our spherical triangles 
have always each of their sides less than a semicircumfer- 
ence; from which it follows that any one of their angles is 
always less than two right angles. For (see diagram to Prop. 
IV.) if the side AB is less than a semicircumference, and AC 
is so likewise, both those arcs will require to be produced 
before they can meet in D. Now the two angles ABC, CBD 
taken together, are equal to two right angles; hence the angle 
ABC itself, is less than two right angles. j 

We may observe, however, that some spherical triangles do 
exist, in which certain of the sides are greater than a semi- 
circumference, and certain of the angles greater than two right 
angles. ‘Thus, if the side AC is produced so as to form a whole 
circumference ACE, the part which remains after substracting 
the triangle ABC from the hemisphere, is a new triangle also 
designated by ABC, and having AB, BC, AIDC for its sides. 
Here, it is plain, the side AFDC is greater than the semicir- 
cumference AED ; and, at the same time, the angle B oppo- 
site to it exceeds two right angles, by the quantity CBD. 

The triangles whose sides and angles are so large, have been 
excluded from our Definition; but the only reason was, that 

3 


22 SPHERICAL GEOMETRY. 


the solution of them, or the determination of their parts, is al- 
ways reducible to the solution of such triangles as are com- 
prebended by the Definition. Indeed, it is evident enough, that 
if the sides and angles of the triangle ABC are known, it will 
be easy to discover the angles and sides of the triangle which 
bears the same name, and is the difference between a hemis- 
phere and the former triangle. 


PROPOSITION XIX. THEOREM. 


The surface of a lune is to the surface of the sphere, as the 
angle of this lune, is to four right angles, or as the arc which 
measures that angle, is to the circumference. 


Let AMNB be a lune; then will its 
A surface be to the surface of the sphere 
¥ as the angle NCM to four right an- 
gles, or as the arc NM to the circum- 
roe ference of a great circle. 
P ! 

ae Suppose, in the first place, the arc 
nN MN to be the circumference MNPQ 
/ as some one rational number is to ano- 
ther, as 5 to 48, for example. The 
Bs circumference MNPQ being divided 
into 48 equal parts, MN will contain 
5 of them; and if the pole A were joined with the several 
points of division, by as many quadrants, we should in the hem- 
isphere AMNPQ have 48 triangles, all equal, because all their 
parts are equal. Hence the whole sphere must contain 96 of 
those partial triangles, the lune AMBNA will contain 10 of 
them ; hence the lune is to the sphere as 10 is to 96, or as 5 to 

48, in other words, as the arc MN is to the circumference. 

If the arc MN is not commensurable with the circumfer- 
ence, we may still show, by a mode of reasoning frequently 
exemplified already, that in this case also, the lune is to the 
sphere as MN is to the circumference. 


Cor. 1. Two lunes are to each other as their respective an- 


gles. 


Cor. 2. It was shown above (Prop. X VIII. Cor. 2.) that the 
whole surface of the sphere is equal to eight tri-rectangular 
triangles ; hence, if the area for one such triangle is taken for 
unity, the surface of the sphere will be represented by 8. This 
granted, the surface of the lune, whose angle is A, will be ex- 
pressed by 2A (the angle A being always estimated from the 


SPHERICAL GEOMETRY. 23 


right angle assumed as unity :) since 2A: 8:: A: 4. Thus 
we have here two different unities; one for angles, being the 
right angle; the other for surfaces being the tri-rectangular 
spherical triangle, or the triangle whose angles are all right, 
and whose sides are quadrants. 

Or if the area of one such triangle is represented by T, the 
surface of the whole sphere will be expressed by 8T, and the 
surface of the lune whose angle is A, will be expressed by 
2A xT. for 

4:3 A-2 J8T 2A GT 
in which expression, A represents such a part of unity, as the 
angle of the lune is of one right angle. 


Scholium. The spherical ungula, bounded by the planes AMB, 
ANB, is to the whole solid sphere, as the angle A is to four 
right angles. For, the lunes being equal, the spherical ungulas 
will also be equal; hence two spherical ungulas are to each 
other, as the angles formed by the planes which bound them. 


PROPOSITION XX. THEOREM. 


Two symmetrical spherical triangles are equal in surface. 


Let ABC, DEF be two symme- D A 
trical triangles, that is to say, two 
triangles having their sides AB= 
DE, AC=DF, CB=EPFP, and yet in- 
capable of coinciding with each | ? P 
other: we are to show that the sur- -l~ G 
face ABC is equal to the surface 
DEF. 

Let P be the pole of the small cir- ‘ B 
cle passing through the three points A, B,C; from this point 
draw (Prop. V1.) the equal arcs, PA, PB, PC; at the point F, 
a the angle DFQ=ACP, the are FQ=CP; and join DQ, 
The sides DF, FQ are equal to the sides AC, OUP; the an- 
gle DFQ=ACP: hence (Prop. XIL) the two triangles DFQ, 
ACP are equal in all their parts ; hence the side DQ=AP, and 
the angle DQF=APC. 

In the proposed triangles DFE, ABC, the angles DFE, ACB 
opposite to the equal sides DE, AB, being equal (Prop. XIII.) 
if the angles DFQ, ACP, which are equal by construction, be 
taken away from them, there will remain the angle QFE, 
equal to PCB. Also the sides QF, FE are equal to the sides 
PC, CB; hence the two triangles FQE, CPB are equal in 


24 SPHERICAL GEOMETRY. 
all their parts ; hence the side QH=PB, and the angle FQE 
B 


Now, the triangles DFQ, ACP, which have their sides re- 
spectively equal, are at the same time isosceles, and capable of 
coinciding, when applied to each other ; for having placed PA 
on its equal QF, the side PC will fall on its equal QD, and 
thus the two triangles will exactly coincide ; hence they are 
equal; and the surface DQF=APC. For a like reason, the 
surface FQE=CPB, and the surface DQE=APB; hence we 
have DQF+FQE—DQE=APC+CPB—APB, or DFE= 
ABC; hence the two symmetrical triangles ABC, DEF are 


equal in surface. i 


Scholium. The poles P and Q might lie within the triangles 
ABC, DEF: in which case it would be requisite to add. the 
three triangles DQF, FQE, DQE together, in order to make 
up the triangle DEF; and in hike manner to add the three tri- 
angles APC, CPB, APB together, in order to make up the tri- 
angle ABC: in all other respects, the demonstration and the 


result would still be the same. 
PROPOSITION XXI. THEOREM. 


If the circumferences of two great circles intersect each other 
on the surface of a hemisphere, the sum of the opposite trian- 
gles thus formed, is equivalent to the surface of a lune whose 
angle is equal to the angle formed by the circles. 


Let the circumferences AOB, COD, 
Intersect on the hemisphere OACBD ; 
then will the opposite triangles AOC, 
BOD be equal to the lune whose an- 
gle is BOD. 

For, producing the arcs OB, OD 
on. the other hemisphere, till they 
meet in N, the arc OBN will be a se- 
mi-circumference, and AOB one also ; 
and taking OB from both, we shall 
have BN=AO. For a like reason, we have DN=CO, and 
BD=AC. Hence the two triangles AOC, BDN have their 
three sides respectively equal; besides, they are so placed as 
to be symmetrical; hence (Prop. XIX. Sch.) they are equal 
m surface, and the sum of the triangles AOC, BOD is equiva-, 
lent to the lune OBNDO, whose angle is BOD. 

Scholium. It is likewise evident that the two spherical pyra- 
mids, which have the triangles AOC, BOD for bases, are to- 
gether equivalent to the spherical ungula whose angle is BOD. 


SPHERICAL GEOMETRY. ™ 25 
PROPOSITION XXII. THEOREM. 


The surface of a spherical triangle is measured by the excess 
of the sum of its three angles above two right angles, multi- 
plied by the tri-rectangular triangle. 


Let ABC be the proposed triangle : 
produce its sides till they meet the 
great circle DEFG, drawn at plea- 
sure without the triangle. By the 
last Theorem, the two triangles ADE, 
AGH, are together equivalent to the 
lune whose anyle is A, and which is 
measured by 2A.T (Prop. XIX. Cor. 
2.) Hence we have ADE+AGH= 
2A.T; and, for a like reason, BGF+ 
BID=2B.T, and CIH+CFE=2C.T. 
But the sum of these six triangles exceeds the hemisphere by 
twice the triangle ABC, and the hemisphere is represented by 
4T ; therefore, twice the triangle ABC is equal to 2A.T+ 
2B.T+2C.T—4T ; and consequently, once ABC=(A+B+C 
—2)T ; hence every spherical triangle is measured by the sum 
of all its angles minus two right angles, multiplied by the tri- 
rectangular triangle. 


Cor. 1. However many right angles there may be in the 
sum of the three angles minus two right angles, just so many 
tri-rectangular triangles, or eighths of the sphere, will the pro- 
posed triangle contain. If the angles, for example, are each 
equal to 4 of a right angle, the three angles will amount to four 
right angles, and the sum of the angles minus two right angles 
will be represented by 4—2, or 2; therefore the surface of 
the triangle will be equal to two tri-rectangular triangles, or to 
the fourth part of the whole surface of the sphere. 


Scholium. While the spherical triangle ABC is compared 
with the tri-rectangular triangle, the spherical pyramid, which 
has ABC for its base, is compared with the tri-rectangular 
pyramid, and a similar proportion is found to subsist between 
them. The solid angle at the vertex of the pyramid, is in like 
manner compared with the solid angle at the vertex of the tri- 
rectangular pyramid. These comparisons are founded on the 
coincidence of the corresponding parts. If the bases of the 
pyramids coincide, the pyramids themselves will evidently co- 
incide, and likewise the solid angles at their vertices. From 
this, some consequences are deduced. 

3% 


- 
— 


26 °° SPHERICAL GEOMETRY. 


First. Two triangular spherical pyramids are to each other 
as. their bases ; and, since a polygonal pyramid may always be 
divided into a certain number of triangular ones, it follows that 
any two spherical pyramids are to each other as the polygons 
which form their bases. 

Second. The solid angles at the vertices of these pyramids 
are also as their bases: hence, for comparing any two solid 
angles, we have merely to place their vertices at the centres 
of two equal spheres, and the solid angles will be to each other 
as the spherical polygons intercepted between their planes 
or faces. 

The vertical angle of the tri-rectangular pyramid is formed 
by three planes, at right angles to each other. This angle, 
which may be called a right solid angle, will serve as a very 
natural unit of measure for all other solid angles. If, for ex- 
ample, the area of the triangle is 2 of the tri-rectangular tri- 
angle, then the corresponding solid angle will also be 2 of the 
right solid angle. 


PROPOSITION XXIII... THEOREM. 


The surface of a spherical polygon is measured by the sum of 
all its angles, minus two right angles multiplied by the num- 
ber of sides in the polygon less two, into the tri-rectangular 
iriangle. 


From one of the vertices A, let di- fi 

agonals AC, AD, be drawn to all the 

other vertices; the polygon ABCDE D 
will be divided into as many triangles, 

minus two, as it has sides. But the 

surface of each triangle is measured by f 
the sum of all its angles minus two 
right angles, into the tri-rectangular tri- 
angle; and the sum of the angles in all A 
the triangles is evidently the same as that of all the angles of 
the polygon: hence, the surface of the polygon is equal to the 
sum of all its angles, diminished by twice as many right angles 
as it has sides, less two, into the tri-rectangular triangle. 


Scholium. Let s be the sum of all the angles in a spherical 
polygon, n the number of its sides, and T the tri-rectangular 
triangle ; the right angle being taken for unity, the surface of 
the polygon will be measured by 


| (s—2 (n—2,)) T, or (s—2 n+4) T 


ANALYTICAL PLANE TRIGONOMETRY. 


CHAPTER I. 


Piane Trigonometry is the science which treats of the re- 
lations of the sides and angles of plane triangles. 

In every triangle there are six parts: three sides and three 
angles; which have such relations to each other that the value 
of one depends on the value of the others; and if a sufficient 
number of these are known the others may thereby be deter- 
mined. 

~The sides of triangles consist of absolute magnitude, but the 
angles are only the relations of those sides to each other in 
position or direction, without regard to their magnitudes. 

Angles have no absolute measure in terms of the sides; but 
are, nevertheless, susceptible of measure ; forif two lines meet 
each other the space included between them within a given 
distance from their point of contact is proportional to their 
mutual inclination, and hence (Prop. XVIII. Cor. B. III. Eu. 
Geom.) the arc of the circumference of a circle intercepted by 
two lines drawn from its centre, may be regarded as the mea- 
sure of the angle or inclination of those lines, and therefore the 
arc of the circumference may be regarded as the measure of 
angular magnitude. 

For this purpose the circumference of the circle is supposed 
to be divided into 360 equal parts, called degrees, and each of 
those degrees is divided into 60 equal parts called minutes, and 
each minute into 60 equal parts called seconds ; and so on, to 
thirds, fourths, &c. 

These divisions are designated by the following characters, 
orut &ec. Thus the expression 380° 20’ 12” 22”, repre- 
sents an arc or an angle of 30 degrees 20 minutes 12 seconds 
22 thirds. 

The circumference of any circle may in this manner be ap- 
plied as the measure of angles, without regard to its magni- 
tude or the length of its radius; hence a degree is not a mag- 


28 ANALYTICAL PLANE TRIGONOMETRY. 


nitude of any definite length, but is a certain portion of the 
whole circumference of any circle, for it is evident that the 
360th part of the circumference of a large circle is greater 
than the same part of a smaller one, but the number of de- 
grees in the small circumference is the same as in the large one. 

The fourth part of the circumference of a circle is called 
a quadrant and contains 90.degrees: hence 90 degrees is the 
measure of the right angle. 


Thus, if we draw two straight lines 
AD, BE, so as to cross each other at 
right angles, and from their point of inter- 
section, C, we discribe a circle with any 
radius so as to cut those lines in any 4A 
points, as a, b, d, e, the circumference +t 
the circle will thus be divided into four 
equal arcs, ab, bd, de, ea, each of which 
measures or subtends a right angle at 
the centre C, of the circle. 

If a line CP be made to revolve round a fixed point C as 
the centre of a circle, and so as to pass successively through 
every point of the circumference, commencing in the point a, 
then, while it is in the position Ca, or while it coincides with 
the line Ca, those two lines form but one, and intercept no are 
on the circumference of the circle, and hence form no angle 
with each other ; but when the line CaP 
comes into the position CP, it forms with 
AC an acute angle at C, which is mea- 
sured by the arc aP,and when it comes 
into the position COP, it then forms a right 
angle ACP with the line AC, which angle 
is measured by the quadrant ab. Now 
let it come into the position CP,, and the 
angle which it forms with CA, will be 
measured by the arc aP,, which is greater than a right angle, 
and hence is an obtuse angle. 

Let it now come into the position CdP ; it then coincides 
with the right line Cd, which is a portion of the line AC pro- 
duced, since the line CP, in this position, coincides with the line 
AD, it can be said to form with it no angle; yet the space 
passed over by the line CP, from the position CaP, is equal to 
two quadrants, or two right angles equal to 180 degrees, and 
for trigonometrical investigation the lines CoP and CA are 
said to subtend the angle measured by the arc abd. 

Alter passing the point d, and coming into the position CP,, 
it forms with AC, and on the upper side of it the angle P,CA 


We -* 


~ 
a 


ANALYTICAL PLANE TRIGONOMETRY. 29 


measured by the arc aeP,, but having passed over the arc 
abdP,, is said to contain, with the line GA, the angle ACP, on 
the upper side of those lines measured by the are abdP.,, 
‘greater than two right angles. When it comes in the posi- 
tion CeP it is said to subtend, with the line AC from the same 
side of it, the angle measured by the arc abde, or three quad- 
rants, equal to three right angles. 

When in the position CP,, it is said to contain with CA, and 
on the same side of it, an angle greater than three right angles. 

Finally, when the line CP’ has completed an entire revolu- 
tion, having returned to its original position, CA, it will have 
formed an angle with it equal to four right angles. 

If the line CP continues to revolve, it is manifest that the 
angle will increase, and may with this view form with CA, 
angles greater than four, than five, or than any given number 
of right angles. 

ab is called the first quadrant of the circle, bd the second, 
de the third, and ea the fourth quadrant. 

It must be borne in mind, that the line CP cannot, geome. 
trically, be said to contain with another line, AC, an angle 
greater, nor quite equal to, two right angles, but in view of its 
supposed motion round one of its extremities, C, as a centre, 
it is said to contain, with the line AC, all the angular space 
through which it has passed in its revolution. 

Thus, let the line CP have performed one complete revolu- 
tion, from the position CaP to the same position again, the 
angle which it forms with the line CA, though absolutely no- 
thing, is in view of its supposed motion measured by the quad- 
rants ab+bd+de-+ea, each of which quadrants are readily re- 
cognized as being contained by their several lines of division, 
when by removing those lines of division of the circumference, 
those several angles are all converted into one containing the 
whole circumference ; hence, in view of this motion or rela- 
tion of the two lines, they are said to contain an angle mea- 
sured by the whole circumference, 


30 ANALYTICAL PLANE TRIGONOMETRY. 


DEFINITIONS AND ILLUSTRATIONS. 


The following symbols are sometimes used. 

1. The complement of an arc or of an angle, is what re- 
mains after taking that arc or that angle from 90 degrees. 
Thus, if 6 be any arc or angle, the complement is 90°—§ 

2. Supplement of an arc or an angle, is what remains after 
taking that arc or angle from two right angles, or 180 de- 
grees. Thus, if 4 be any arc or angle, the supplement of é is 
180°—4. 

If AP be any arc, and ACP be any angle, 
é measured by that arc, then the complement 
of the angle 4 is the angle PCB measured by 
the arc PB, and the supplement of the angle 4 
is the angle PCD measured by the arc PBD, 
and if BCP is any angle é measured by the 
arc PB, then PCA is the complement of 4, and if PCD is any 
angle 4, then will PCA be its supplement. 

3. To represent the ratios of the sides and angles of trian- 
gles, right lines are drawn in and about a circle called sines, 
tangents, secants, &c. 

Draw two right lines AD,BE cutting each 
other at right angles in the point C, with the 
centre C and any distance as radius, describe a 
circle cutting the lines in the points A, B, D, E. 

Draw the radius CP forming with CA any 
angle ACP=é. From P draw PS perpendi- 
cular on CA. From A draw AT a tangent 
to the circumference at A. Produce CP to 
meet AT in T. 

4. Then the ratio of PS to the radius CA of the circle, is 
called the sine of the angle PCA. 

Or Lashes b. 
CA 

5. The ratio of AT to the radius CA of the circle, is called 

the tangent of the angle TCA or PCA. 


Hence, AT _ tan. 6. 


6. And the ratio of CT to the radius is the secant of the 
angle PCA, 


CT 
Or, ey a 6. 


ANALYTICAL PLANE TRIGONOMETRY. 31 


7. The ratio of AS to the radius of the circle, is called the 
versed sine of the angle PCA. nasal 
Or, a sin. 6, 
8. The sine of the complement of an angle, is called the sine 
complement, or cosine of that angle. 
Thus, sin. (90°—%)=cos. 6, hence cos. (90°——4)=sin. 4. 
9. The tangent of the complement of any given angle, is 
called the cotangent of that angle, 
Thus, tan. (90°—4)=cot. ¢, hence cot. (90°—#)=tan. 4 
10. The secant of the complement of any given angle, is 
called the cosecant of that angle. 
Or, sec. (90°—#)=cosec. 4, hence cosec. (90°—4)=sec. 4. 
11. The versed sine of the complement of any angle, is call- 
ed the co-versed sine of that angle. 
Or, v. sin. (90°—4)=co-v. sin. 4, 
and hence co-v. sin. (90——4)=v. sin. 4. ; 
In order to show that the ratio of CS to the radius of the 
circle in the last figure, is the cosine of the angle PCA; that 
is, the sine of its complement, 


v 


Or that E> =cos. 4, 


be 
Draw a circle A’/B’D’E’ equal to the circle ‘2 


ABDE, and from C’ the centre, draw C’P’, Aly 
making with C’A’ the angle P’C’A’ equal to 4: 


D’ 
the angle PCB; that is, to the compliment 
of PCA, or to (90°—4). 


E 


Then, since CP is equal to C’P’, and the 
angles at S and 8’ are right angles, the angle 
CPS equal to the angle P’C’S’ the two trian- A 
gles PCS, P’C’S’ are equal in every respect ; 
fo—C'S’, CS=P's 


CS P's’ 
CA CA 

=sin. P’C’A’ by Def. 

=sin. (90°—4) by construct. 

=cos. 4 by Def. 

We have hitherto considered an angle PCA less than a right 

angle, but the same definitions are applied, whatever may be 
the magnitude of the angle. 


Therefore, 


+ 


32 ANALYTICAL PLANE TRIGONOMETRY. 


Thus, for example, let us take an angle 
P,CA situated in the second quadrant, that ts, 
an angle greater than one right angle, and 
less than two. 

From P, let fall P,S, perpendicular on 
AD, from D draw DT, a tangent to the cir- 
cle at D, meeting CP, produced i in T, ; then, 
as before, 


PBs sin, BCA 

GA 

CS, 

GA 2—cos. P,CA 

DY: 

“dl tan. Wg CA 

ad be 

CA 2—see. PCA 
; AS, 

CA 


Again, let the angle in question be situated in the third quad- 
rant, that is, let it be an angle greater than two, and less than 
three right angles. 

Making a construction analogous to that in the two former 
cases, we shall have 


3° sin P,CA 
Hs cos, Ba. 
= otan, FAGSs: 
Sey P,AC 
sae sin. P,CA. 


Lastly, let the angle be situated in the fourth quadrant; that 
is, let it be an angle greater than three, and less than four 
right angles, then as before, 


few 


ANALYTICAL PLANE TRIGONOMETRY. 33 
CT 
GA ee P,CA 
Se sin. P,CA. 


We shall now proceed to establish some important general 
relations, between the trigonometrical quantities which are 
immediately deducible from the above definitions, and from 
the principles of Geometry. 


T 


Resuming the figure of Def. (8): 
Since CSP is a right-angled triangle, and 
- CP the hypothenuse, 
PS’+CS?=CP* 
Dividing by CP”, 


PS? CS 1 
CPT CP?” 
that is, sin.%6+cos.d=1 - - - - - ew + hl) 
The triangles PSC, TAC, are Peeangulse and similar ; 
hence, 
PS_AT 
CS “CA 
PS 
CA AT 
Therefore cs’ CA 
CA 
that is, ona Panta er oe oe a) 
cos. 6 
In last case, for 4 substitute (90°—4) ; then 
sin. (90°—4) _ o 4 
cos. (90°—4) ae 
Or, S08de | ile = Rovdoe HieW i= Thoth SSS) 
sin. 8 
From (2) and (3) we have 
sin. é oh F 
os.6 at 
Hence, tan. 6= a or, tan. 6cot.d=1- + - - - - (4) 
By similar triangles ‘CTA, CPS. 
igs cP 
CA CS 
os 
=CS 
I 
4 


34 ANALYTICAL PLANE TRIGONOMETRY. 


i 
Or, sec. sieaaee 7 OF; Sec. écos.6=1 - - (5) 


By Definition, 


cosec. 6=sec. (90°—#) 


l f 4 
= t case & 
cos. (90°28) by the last case, es 
] 


. ‘_. « 
“ 


= ste or, cosec. sin. é=1 - - (6) 
Since CAT is a right angled triangle, and CT the hypothe- 
nuse 
CA°+AT°=CT? — a 
Dividing by CA’, 
sale ba bo ad Wi 
GAry CAS 
that Is, 1--tan” d=sec.27) = Poker ate ee 
By (8) we have a ually j 
sin. 6 
Sig's os.” 4 
Therefore, cot.” (= 
Adding 1 to each side of the equation, 
wp : lcomee ees 
¥. sin. r 
aa _sin.’ 6+cos.* 4 
mae ; 
r Tae aj by (1) 
= COseC.: O)Dy (0) ae > =: ou ee) 
By Definition, abana 5" 
versin. FeV ae ® 
_CA—CS 
= Cider 
ls CS 
™? CA 
=I—cos.6- - - - - - = (9) 


By Definition, 


coversin. 6=versin. (90°—4 
= 1—cos.(90°—4,) a the last case. 
<= ]—dgin) dc 12) eoigiesiod ollie 1h 


The above results, which are of the highest importance in 
all trigonometrical investigations, are collected and arranged 
in the following table, which ought to be committed to memo- 


ya 


ANALYTICAL PLANE TRIGONOMETRY. 35 


TABLE I. 
1.- Sin. 6+-cos.? 6=1 
sin. 6 
Zt ee tania 
or cos. 4 
fa cos. 4 
eas 3. Binh =cot. 4 
42 *an:6 Coe ei 
5. seciidéosre Sti 
6. cosec. 4 sin. 6 =1 
# 7. %I+tan.? 4 =sec.? 6 
8. 1+cot.’ 4 =cosec.” 6 | 
9. v. sin. 4 = }—cos, 6 
10. coversin. 6 a] sine 8! 


12. The chord of an arc is the ratio of the straight line join- 
ing the two extremities of the arc to the radius of the circle. 


PROPOSITION. 
The chord of any arc is equal to twice the sine of half the are. 


Take any arc AQ, subtending at the 
centre of the circle the angle ACQ=4. 

Draw the straight line CP bisecting the 
angle ACQ. 

Join A, Q; from P let fall PS perpen- 
dicular on CA. 

Since CP bisects ACQ, the vertical an- 
gle of the isosceles triangle ACQ, it bi- 
sects the base AQ at right angles. 

Therefore, AO=OQ, and the angles at O are right angles. 

Again, since the triangles AOC, PSC, have the angles CSP, 
COA, right angles, and the angle PCS common to the two 
triangles, and also the side CP of the one equal the side CA 
of the other, these triangles are in every respect equal. 


. PS =AO=0Q 


ADF, A emtd tthe: 
“AQ _,PS 
"* CA CA 
or, chord 6 =2 sin. PCA 
6 
ae) SH 5 


We shall now proceed to explain the principle by which the 
signs of the trigonometrical quantities are regulated — 


* 


od 


36 ANALYTICAL PLANE TRIGONOMETRY. 


All lines measured from the point C along B 
CA, that is, to the left, are considered pos- 
itive, or have the sine +. 

All lines measured from the point C 


along CD, that is, in the opposite direction « = D 
to the right are considered negative, or have 
the sign —. 
All lines measured from the point C along 
CB, that is, upwards, are considered posi- E 


tive or have sign +. 

All lines measured from the point C along CE, that is, in 
the opposite direction downwards, are considered negative, or 
have the sign —. 

Let us determine according to this principle, the signs of the 
sines and cosines of angles in the different quadrants. 


B 
In the first quadrant, sin. I=GA 
CS 


Here PS=Cc is reckoned from C along 
CB upwards, and is therefore positive. 

CS is reckoned from C along CA, to the 
left and is therefore positive. 

In the first quadrant, therefore the sine 
and cosine are both positive. 


: sy 

In the second quadrant, sin. 6= GA 
CS 

y= 2 

cos.f= 


Here P,S,=Cc, is reckoned from C 
along CB upwards, and is therefore posi- 
tive. 

CS, is reckoned from C along CD to 
the right and is therefore negative. 

In the second quadrant, therefore, the sine 
is positive and the cosine negative. 


: P95 

In the third quadrant, sin. 6= CA 
CS 

cos, 6=_—_3 
CA 


Here P,S,=Cc, is reckoned from C 
along CE, downwards, and is therefore 
negative. 

CS, is reckoned from C along CD, to 
the right, and is therefore negative. 


. 


Me 


ANANLYTICAL PLANE TRIGONOMETRY. 37 


In the third quadrant, therefore, the sine 
‘and cosine are both negative. 


BS 


4 


In the fourth quadrant, sin. ere 


cos. jun Oa 


CA A D 
Here P,S,=Cc, is reckoned from C along \<o a 
CE downwards and is therefore negative. — Nes 
CS, is reckoned from C along CA, to the E 
left and is therefore positive. 

In the fourth quadrant, therefore, the sine is negative, and 
the cosine positive. 

Hence we conclude, that the sine is positive in the first and 
second quadrants, and negative in the third and fourth ; and the 
cosine is positive in the first and fourth, and negative in the 
second and third, or in other words: 

The sine of an angle less than 180° is positive and the sine 

- of anangle greater than 180° and less than 360° is negative. 

The cosine of an angle less than 90° ts positive, the cosine of 
an angle greater than 90°, and less than 270°, is negative, and 
the cosine of an angle greater than 270°, and less than 360,° is 
positive. 

The signs of the sine and the cosine being determined, the 
signs of all the other trigonometrical quantities may be at once 
established by referr:ng to the relations in Table 1. 

Thus, for the tangent, . 
sin. 4 
; Uae Oe COs, 6 
Hence, it appears that when the sine and cosine have the 
same sign the tangent will be positive, and when they have 
different signs it will be negative. 
Therefore, the tangent 1s positive in the first and third quad- 
rants and negative in the second and fourth. 
The same holds good for the cotangent; for 


cos. 6 
cot. 6=-— 
sin. 6 
Again, since Laas bl 
cos. 4 


the sign of the secant is always the same with that of the co- 


sine ; and, since 
cosec. = 


sin. 4 
in like manner, the sign of the cosecant is always the same with 
that of the sine. 


The versed sine is always positive, being reckoned from A 
always in the same direction. 


38 ANALYTICAL PLANE TRIGONOMETRY. 


It is sometimes convenient to give differ- P 
ent signs to angles themselves. We have ce 
hitherto supposed angles of different magni- 
tudes to be generated by the revolution of A aN 
the moveable radius CP round C in a direc- 
tion from left to right; and the angles so 
formed have been considered positive, or af- P’ 
fected with the sign +. If we now suppose the angle 8=é 
to be generated by the revolution of the radius CP’ in the op- 
posite direction, we may, upon a principle analogous to the 
former, consider the angle § as negative, and affect it with the 
sign —. 

“We shall now determine the variations in the magnitude of © 
the sine and cosine for angles of diflerent magnitudes. 


In the first quadrant : 

Let Chat Be, GR. :....... be different 
positions of the revolving radius in the 
first quadrant; and from P, P,, Ps, ........ 
draw PS, P,S,, P,S,, perpendiculars on 
CA. 

It is manifest, that as the angle in- 
creases the sine increases ; for 

Py San. PS aa PES pt eis 
CA Ch & ¢ OR 456A 

When the angle becomes very small, PS becomes very 
small also ; and when the revolving radius coincides with CA, 
that is, when the angle becomes 0, then PS disappears alto- 


gether, and is =0. ” 
Hence since, generally, sin. ‘a. and since, when 6=0, 
PS=0; CA, 
‘ 0 
sin. FON 


On the other hand. when the angle becomes equal to 90°, PS 
coincides with CB, and is equal to it. 


Hence since, generally, sin. i= and since, when §=90°, 


PS=CB; 


1 ae @B=CA. . 
_ Again, it is manifest, that as the angle increases the cosine 
diminishes ; for 
CS CS, ed b= eel Oe 
CA- CA CA CA 


ANALYTICAL PLANE TRIGONOMETRY. 39 


- 


When the angle is very small, CS is very nearly equal to 


~ CA; and when the revolving radius coincides with CA, that 


is, when the angle is 0, then CS coincides with CA and is 
equal to Me ° 
Hence since, generally, cos. = 77; GA? and since, when 6=0, 
CS=CA; 
% “COS. gaa si 
ameter 
On the other hand, as the angle increases, CS diminishes, 
and when the angle “eR equal to 90°, CS disappears al- 
together, and is =0. 


Hence since, generally, cos. oor and since, when 6=90°,CS 
=0; 
| « -@ 
. COS. 90°= ay 5s Cae 
Let us now take different positions of the revolving radius 
in the second quadrant. 
It is manifest, that as the angle in- 
creases the sine diminishes ; for 
» PSPor Se sail Foyer oe 
CA CA > CA 
As the angle goes on increasing, PS 
goes on diminishing ; and when CP coin- 
cides with CD, that is, when the angle 
becomes equal to 180°, PS pea al- 
together and is equal 0. 


PS 
Wende since, gencrally, sin. t= Tai and since, when §=180°, 
PS=0; 
-, sin. 180°=0. 
On the other hand, as the angle increases the cosine increases ; 
for 


O87 Gg CSPRECS! 
CA CA <CA 


‘and when the revolving radius coincides with CD and the an- 


gle becomes 180°, CP coincides with CD and is equal to it. 


1 


Hence since, generally, cos. saat OF and since, when §6=180°, 


CS=CD; 
iso 1 
cos iA a 
. CD=CA. 


The negative sign Hite is employed, because the cosine is 
reckoned to the right along CD. 


#5 


“40 ANALYTICAL PLANE TRIGONOMETRY. 


Reasoning in the same manner for the third and fourth 


quadrants, we shall find, ' 
sin. 370° =} 
* eosae70°=—""0 
sin. 360°= 0 


cos. 386U0°= 1. 
Thus, it appears, 
That as the angle increases in the first quadrant, from 0 up to 
90°53 
The sine, being positive, increases from 0 up to 1, 
The cosine being positive decreases from 1 down to 0. 
That, as the angle increases in the second quadrant, from 90° 
up to 180°, 
The sine, being positive, decreases from 1 down to 0, 
The cosine, being negative, passes from 0 to —1. 
That, as the angle increases in the third quadrant, from 180° 
up to 270°, 
The sine, being negative, passes from 0 to —1, 
The cosine, being negative, passes from —1 to 0. 
That, as the angle increases in the fourth quadrant from 270° 
up to 360°, 
The sine, being negative, passes from —1 to 0, 
The cosine, being positive, increases from 0 up to 1. 


The variations in the magnitude of the sines and cosines, 
being those of the other trigonometrical quantities may be de- 
termined by the means of the relations in Table I. 


sin. 4 
Thus, since, tan. 6=——— 
cos. 4 
; Prati NAD iy 
aD y | cos.0hel p 
op, 810790" LL 
tan. 90°=— 590° 0 ad infinitum, or @. 


The truth of this last relation may be readily illustrated, by 
referring to the geometrical construction; when it will be 
seen that for the angle of 90° AT becomes parallel to CP; 
and therefore, the point T, in which the two lines meet, is at 
an infinite distance. 

_ So, also, cot. 0=@ 

. cot. 90°=0 

and so for all the rest. 


We shall next proceed to point out some important general 
relations, which exist between the trigonometrical functions of 
angles less than 90° and those of angles greater than 90° — 


ANALYTICAL PLANE TRIGONOMETRY. 41” 


_ Draw CP, making with CA any angle 
~~ PCA which we may call 4; let fall PS 
perpendicular from P on CA. Draw 
CP’, making with BC the angle BCP’— 
PCA=4; and from P’ let fall P’S’ per- 
pendicular on CD. 
Then the angle P’-CA—90°+4, 
The two triangles PCS, P’CS’, have 
the side PC of the one equal to the side 
P’C of the other, also the angles at S and 
S’ right angles, and the angle CPS of the one equal to the an- 
gle P’CS’ of the other; therefore the two triangles are in 
every respect equal ; and 


BS=CS’', .CS=P'S’ 


P'SL.. CS 
Therefore, CA CA 
Or, sin. P’‘CA=cos. PCA, 
that is, sin. (90°+4)=cos. 4, 
Again, 
| CS’ PS 

GARLCA 
Or, —cos. P/CA=sin. PCA, 
that is cos. (90°+6)=——sin. 4. 


6 with CA, and draw CP’, making with 
CD the angle P’CD, equal to 4. 


Then the angle P’/CA=180°—4. | 
The two triangles PCS, P’CS’ are ma- fee el D 
nifestly in all respects equal; and 
PS=P'S”CS=En" 
PS _Ps' 
CA CA 

that is, sin. é=sin. (180°—#) 
an important proposition which enunciated in words, is, the 
sine of an angle is equal to the sine of its supplement. 

Again, 


As before, draw CP, making any angle e 
P 


Therefore, E 


CS_ Cs’ 
CA CA 
cos. 46=—cos. (180°—4), 
that is, the cosine of an angle, and the cosine of its supplement 
are equal in absolute magnitude, but have opposite signs. 


4 


42 ANALYTICAL PLANE TRIGONOMETRY. 


B 


If, as in the annexed figure, we draw 
CP’, making with CD an angle DCP’ 
equal to the angle 4, we shall find in like 
manner, 

sin. (180°-+¢6)=—-sin. 8 
cos. (180°+0)=cos. (180°—#) 
=—cos. 4. 


If we draw CP’, making with CE an 
angle ECP’=4, then 
sin. (270°—6)=—cos. 6 
cos. (270°——4) =— sin. 4 
as is evident from Def. 8, and the rule for 
signs; and, in like manner, we may pro- 
ceed for angles in the fourth quadrant. 
These relations being established between 
the sines and cosines, the corresponding 
relations between other trigonometrical functions may be de- 
duced immediately from Table 1. 
Thus, 3 ASL XID SRO) 
tan. (O00) east (90° <8} 


cos. 4 


esi 8 

ee rh 
sin. (180°—4) 
tan. (486°—3)—- pa en. EV: 
( ) cos. (180°—4) 

sin. 6 


~ aos, d 
romper ae 
and so for all the rest 


The student may exercise himself by verifying such of the 
results in the following table as have not been formally de- 
monstrated. 


ANALYTICAL PLANE TRIGONOMETRY. 43 


*sin. O 
*cos. 0 
*tan. 0 
*cot. O 
sec. 0 
cosec. 0 
*sin. (90°—6) 


*cos. (90°——8) 
*tan. (90°—4) 


*cot. (90°—4) 
sec. (90°—4) 


cosec. (90—4) 


*sin. 90° 
*cos. 90° 
*tan. 90° 
*cot. 90° 
sec. 90° 
cosec. 90° 
*sin. (90°+64) 
*cos. (90°+8) _ 
*tan. (90° +6) 
acot. (90°-+2) 


sec. (90°+4) * 
_cosec. (90°+68) 
*sin. (180°—48) 
*cos (180°—4) 
*tan. (180°—8) 
*cot. (18U°—4) 

sec. (180°——) 

cosec. (180°—4) 


* sin. 180° 
*cos. 180° 
*tan. 180° 
*cot. 180° 
sec. 180° 
cosec. 180° 


TABLE IL. 
=—0 *sin. (180°+6) =—sin. 6 
=! *cos. (180°+4) . =—cos. 4 
=0 tan. (180°+-6)  =tan..4 
= @ cot. (180°+4)  =cot. 4 
== sec. (180°+6) =—sec. 4 
=’ 2 cosec. (180°+46) =—cosec. 4 
=cos. 4 sin. (270°—-4)  =—cos. é 
=sin. 4 cos. (270°—4) =—~sin. 4 
=cot. 6 tan. (270°—-6) =cot. 4 
=tan. 4 cot. (270°—4) =tan. 4 
=cosec. 4 sec. (270°—#) =—cosec. 4 
= S6Gau cosec. (270°——4) =— sec. 4 
= sin. 270° =—l1 
==) cos. 270° = 
ane e s tan. 270° — Om 
=0 cot. 270° = 
an tis sec. 2 /0° == Gp 
= cosec. 270° =— 
=€05.00 sin. (270°-+4) » =— cos. 4 
= —sin. 4 SOSH 210? 1-0" = sink 
= —cot. 4 tan. (270°+4) . =—cot.é 
= —tan. 8- | cot.(270°+4) ~=—+tan. é 
= —cosec. 4 sec. (270°+6) =cosec. 4 
=sec. 6 cosec. (270°+6) =—sec. 4 
=sin. 6 sin. (360°—0) =—sin. 4 
=—cos. 4 | cos. (360°—é) » =cos. 4. 
=—tan. 4 | tan. (360°—-4) =—tan. é 
= — cot. 8 cot. (3860°—4) =——cot. é 
= —sec. é sec. (860°—0) =sec. 4 
==Ccosec cosec. (360—-4) =—cosec. 4 
= sin. 360° ==(0) 
= —] cos. 360° sor 
—(0 tan. 360° =3() 
=—@® cot. 360° == 
=—] sec. 360° aH 
=o cosec. 360° =@ 


The results in the above table which are most frequently 
used, are marked with an asterisk, and ought to be committed 


to memory. 


We have in the preceding pages confined ourselves to the 
consideration of angles not greater than 360°, but the student 
can find no difficulty in applying the above principles to an- 
gles of any magnitude whatsoever. 


- 


44 ANALYTICA AL PLANE “TRIGONOMETRY. a 
We shall conclude this introductory chapter, by doitbnsirat< 
ing two propositions which are of the highest importance in 
our subsequent investigations. The first is, =~ 
In any right-angled triangle, the ratio which the side oppo- 
site to ne of the acute angles bears to the hypothenuse, is the 
sine of that angle; the ratio which the side adjacent to one of 
the acu e angles bears to the hypothenuse, is the cosine of that 
angle; and the ratio which the side opposite to one of the acute 
angles bears to the side adjacent to that angle, is the tangent of 


that angle. 
Let CSP be a Pane triangle right-angled at S. a 
PS CS PS 
Then, Gp =sin, 'G; Gp aco: C, » ga tan. CG; 
CS SP SC 
or Gp sin te » Gp— Cos: Pp, »Spatan. P 


From C as a centre with the radius CP, 
describe a circle. 

Produce CS, to meet the circumference — 
in A. 

From A draw AT a tangent to the circle , 
at A. ‘ 

Produce CP to meet AT in T. 

Then, from Definitions (2) (8) (4), 


PS. 34 CS Mg 
Gp sin. C, Gp cos Gs Gp tan, 
for CP—AC. 
But the triangles TAC, PSO, are similar ; 
Therefore, AT_PSs =-tan. C, 
CPCS 
Cor. BS —CP-sin.:C=CP cos. P 
CS=CPcossG=—CP sin:-P | 
Pe PS=CS tan. C=CS cot. P Be 


“Mt 
The second proposition is, 
In any plane triangle, the ratio of any two of the sides is 
equal to the ratio of the sines of the angles opposite to them. 
Let ABC be a plane triangle; it is required Cc 


to prove that . 
CB_sin. A CB_ sin. A CA_sin. B # 
CA sin. BBA sin. C’ BA sin. C’ > : 

From C let fall the perpendicular CD on AB. 

Then, since CDB is a plane triangle right-an- D 

gled at D, by last proposition , 

a * 


“4 < 
¥. 


ANALYTICAL PLANE TRIGONOMETRY. A5 


CP—CE dip th +5 ape) nice (Pe) hak” iQ) 
Again, since CDA is a plane triangle right-angle at A, 
; A. SHBG een eo ee ah - - (2) 


Equating these two values of CD, 
CB sin 7 api sin. A; 
C sin. A. 
Therefore, a 
In like manner, by dropping perpendiculars from B and A 
upon the side AC, CB we can prove, 
CB sin. A, CA_ sin. B, 
BA sin. G, BA” sin. C. 
ay In treating of plane triangles, it is convenient to designate 
% _ the three angles by the capital letters A, B, C, and the sides 
opposite to these angles by the corresponding small letters 
a,b,c. According to this notion, the last proposition will be 
a sinw.A a sin.A 0D sin. B 


[_—i: +!  —_--. we =F 
b sin. B oc sin.C c sin,c 


? 


CHAPTER II. 


GENERAL FORMUL. 


% “~ . ° ; ° ° 
Given the sines and cosines of two angles, to find the sine of 
their sum. 


Let ABC be a plane triangle ; from C let fall C C 
perpendicular on AB, 
Let angle CAB=4, 
and angle CBA=8, 
Then, AB = BD+DA @ 
= BC cos, 8 + AC cos. 4, any De. 8B 
because BDC and ADC are right-angled triangles, 


___ Dividing each member of the equation by AB, 
— BC AC ; 
] = AB °°: B + AB cos. 


a 


Ae 
= —G Goos B +5 F cos. 8, by last Prop. in Chap. I. 


“.sin.C = sin.écos. 8+ sin. 8 cos. 4. 
_ _ But, sinceABC is a plane triangle, 4+8+ C = 180° 
- C= 180°—(6+8) 
sin.C = sin. {180°—(6+ 
i= sin (6+8,) because 180°—(4+/) is the supplement 
of (6+.) 
5 


46 ANALYTICAL PLANE TRIGONOMETRY. 


Hence, sin. (8+4) = sin. 4 cos 6+ sin. 8 cos. 6- ~ - - (a) 
That is, the sine of the sum of two arcs or angles is equal the 
sine of the first multiplied by the cosine of the second, plus the 
sine of the second mulitplied by the cosine of the first. 

This expression, from its great importance, is called the 
fundamental formula of Plane Trigonometry, and nearly the 
whole science may be derived from it. 


Given the sines and cosines of two angles, to find the sine of 
their difference. 


By formula (a). 
sin. (0+) = sin. 4 cos. 8+ sin. 8 cos. 4, 

For 4 substitute 180°—4, the above will become 

sin. §{180°—(8-—-8)} = sin. (180°—4) cos. 8 
+ sin. 8 cos.(180°—) 
But sin.§180°—(é—§)} = sin. (6—8) +. 180°—(é—) is the 
supplement of (8—8.) 

And, sin, (180°—#) = sin. 4, 

And, cos.(180°-—4) = — cos. 4 

Substitute, therefore, these values in the above expression, 
it becomes 

sin. (6—) = sin. 4 cos. 8—sin. 8 cos. 4 - - - - (b) 

That is, the sine of the difference of two arcs or angles, is equal 
the sine of the first X cosine of the second, — the sine of the se- 
cond X cosine of the first. 


Given the sines and cosines of two angles, to find the cosine of 
their sum. 


By formula (a) 
sin. (6+ 8)=sin. 4 cos. 8+ sin. 6 cos. 8, 
For 4 substitute 90°+4, the above will become 
sin. {90°+(6+8)}= sin. (90°+4) cos. 8 
+ sin. 8 cos. (90°-+-4) 
But, sin. $90°+(é+8)}=cos. (46+8) by Table II. 


And, sin. (90°+4) =cos. 4. 
And, cos. (90°+3) =—-sin. 4, 

Substituting, therefore, these values in the above expression, 
it becomes, cos. (+8) =cos. 4 cos. 8—sin. @ sin.8 - - (ce) 


That is, the cosine of the sum of two arcs or angles, is equal to 
ihe cosine of the first multiplied by the cosine of the second, mi- 
nus the sine of the first multiplied by the sine of the second. 


. 


ANALYTICAL PLANE TRIGONOMETRY. 47 


Given the sines and cosines of two angles, to find the cosine of 
their difference. 


By formula (@:) 
sin. (6+) =sin. 6 cos. 8+ sin. 8 cos. 4, 
For 4 substitute 90°—4, the above will become 
sin. ${90°—(é—8)}= — sin. (90°—4) cos. 8 
+ sin. 8 cos. (90°—4), 
But, sin. {90°—(é—£)}=cos, (6—8), By Table II. 
sin. (90°—4) cose lth oF) 14.2 
cos. (90°——4) =a) - BS AO Ala: 
Substituting, therefore, these values in the above expression, 
it becomes 
cos. (6—f) =cos. 4 cos. 6+sin.4sin.8 - - (d) 


That is, the cosine of the difference of two arcs or angles, is 
equal to the cosine of the first multiplied by the cosine of the 
second, plus sine of the first into-the sine of the second. 


Given the tangents of two angles, to find the tangent of their sum. 


By Table L.: 
sin. (+8) 
tan. (6+ 8) Smeal ROWS) G+8) 
sin. 6 cos. 8+sin. 8 cos. 4 
~ cos. 6 cos. B—sin. 6 sit. 8 by (@) and (c) 


Dividing both numerator and denominator of fraction by 
cos. 6 cos 8: 


sin. 6 cos. 8 . sin. 8 cos. 4 


cos. cos. 8 cos. cos. 4 
rr sin. 4 sin. 8 
~ eos. 4 cos. 8 
Simplifying, tan. 6+tan. 8 
Tt i—tatedisand syode oll rl) se () 


That is, the tangent of the sum of two ares or angles, is equal 
to the sum of the tangents of the two arcs, divided by 1 minus 
the product of the two tangents. 


Given the tangents of two angles, to find the tangent of their 


difference. 
By Table I.: 
_ sin. (@—8) 
pga See (6 —8) 
sin. 6 cos. 8—sin. 8 cos. 
~ cos. 4 cos. 8+sin. 4 sin. P. by (b) and (d) 


48 ANALYTICAL PLANE TRIGONOMETRY. 


Dividing both numerator and denominator by cos. 4 cos. 8 : 
sin. cos. 8 — sin. 8 cos. 4 
cos.écos.8 cos. 6 cos. B 
ste sin. @ sin. 8 
; cos. 4 cos. 8 
Simplifying, tan. 6— tan. 8 
b+ tan. étane te Ok Lec Oe (7) 


Hence, the tangent of the difference of two arcs or angles, is 
equal to the difference of the tangents of the two arcs, divided 
by 1 plus the product of the two tangents. 


The student will have no difficulty in deducing the following : 
cot. 6 cot. P—1 


cot. (6+5)= ——_——_— 
G+8) cot. P+cot. 6 
cot. 6 cot. 8+1 
(oe 0) ee 
cot. ()—8)= cot. 8—cot. 6 
sec. 4 sec. 8 cosec. 4 cosec. 
SEC. riley sal adi | 
cosec. 4 cosec. S—sec. 4 sec. 8 
sec. sec. 4 sec. 8 cosec. 4 cosec. B 
sec. (6—8)= 


cosec. 4 cosec. 6+sec. 4 sec. 8 
sec. § sec. 6 cosec. 4 cosec. 8 
sec. 6 cosec. 8+sec. 8 cosec.4 
sec. 6 sec. 8 cosec. 6 cosec. B 
sec. 6 cosec. B—sec. 8 cosec. 6 


cosec. (6+8)= 


cosec. (6—P)= 


To determine the sine of twice a given angle. 


By formula (qa) : 
sin. (6+8)=sin. § cos. 8+sin. 8 cos. 6. 
Let 6=8, then the above becomes 
sin. 2 6=sin. 4 cos. é+sin. 4 cos. 4 
=2 sin. 6 cus.6- --- -) -. =, (gl) 
That is, the sine of twice a given angle, is equal to twice the 
sine of the given angle multiplied by its cosine. 


6 
In the last formula, for 4 substitute >; then, 


‘ 6 sus 6 6 
sin. 2X o= Sin. “5 COS. > 
6 ) 
Or, sin. d= 2 sin. —coss—= - - - - = + (g2) 


2 2 


ANANLYTICAL PLANE TRIGONOMETRY. 49 


To determine the cosine of twice a given angle. 


By formula (c) : 
cos. (4+8)=cos. 4 cos. 8 —sin. 4 sin. 8 
Let 6=§8 then the above becomes 


cos. 26=cos.*6—sin2 6 - - 2 + - - (hl) 
By table I. sin.? 6 =1—cos.’ 6; substituting this for sin.* 4: 

COs. = 2 cos.’ §—— 1 oy OU bh 2 (h2) 
Again, since cos.” 6=i— sin.’ 4, substitute this for CUS. 2 

cos. 2 =I. — 2 sin.’ 4 - ie = = - - = (h8) 


Hence, the cosine of twice a given arc or angle, is equal to 1 
minus twice the square of the sine of the given angle. 


To determine the tangent of twice a given angle. 


By formula (e) : 


tan. 6+tan. 8 
ets C18) = —tan.é tan. 8 
Let 6=8, the ate becomes 
2 tan. 4 2 
tan. 2 ace My PSY ekywe bond ke Gtcuetieettiin. Wao (2) 


The tangent of twice a given arc, is equal to twice the tangent 
of the given arc, divided by 1 minus the square of the tangent. 


The student will easily deduce the following : 
cot.*§—1 cot. 6—tan. 4 


cot.2 c= 2cot.4 | 2 
sec.” 6 cosec.” 4 5 
sec. 2 Oe cogent 6— sec’ 4 
sec.” 6 cosec.? 6 sec. 4 cosec.. 4 
cosec. 2 = Shrrentne teh 


2 sec. 6 cosec. 6 ~ 2 


To determine the sine of half a given angle. 
By formula (23) : r 
cos. 24=1—2sin26 °” 


6 
For ¢ substitute >: the above becomes, 


: 1—2 et 
cos. 25°= — 2 sin. 5 
i 4 
Or, cos. 6=1—2 sin.’ > 
a eS ee 6 
‘2 
4 


+ 


50 ANALYTICAL PLANE TRIGONOMETRY. 


That is, the sine of half a given angle or arc is equal to the 
square root of 1 minus the cosine of the arc divided by two. 


Lo determine the cosine of half a given angle. 


By formal (h2): 


Cosi Ribil2ae? D cogti=—] 


. 6 
For 64 substitute zi the above becomes, 


cos, 2 a = 2 cos. ‘a —l 
2 2 
Or, cos/e) == sz ees, eat 
os cos." = 1+ cos. 4 
4 LA set 
C08 g, win Mata ee 


Fo determine the tangent of half a given angle. 
Divide formula (4) by (A): 
6 


12g flaca 
Botta Ice 


acy il 1+cos. 6. . 
ball oe 
8 1—cos é 
Or tans =) te oe he ee 7s 
; 2 1-++cos. 6 ( ) 


Multiply both numerator and denominator by Vi—cos. 6; the 
above becomes, 
1—ceos. 6 


6 
ta at it, = : nt a sae = o ee - - ® 
i 2 sin. 6 (72) 


Multiply both numerator and denominator of (71) by 
V1-+cos. 6; we have, 


sin. 6 
tan. mY = 1chca a ha ee, Cee ( 3) 
The student will easily deduce the following : 
6 Nal 1+cos. 4 
cot. oy = mee 
1—cos. & 
Le 1+ cos. 4 
a sin. 6 
sin. 6 


ANALYTICAL PLANE TRIGONOMETRY. wt 


sec. 4 LR Bice sec. 6 

2 sec. 6+1 

PER ld ps a / 28002 8 6b 
2: sec. 6—] 


To determine the sine of (n+1) 4, in terms of n 6, (n—1) 4 
and 4, 


By formula (a) and (4): 
sin. (8+4)= sin. 8 cos. 46+ sin. 6 cos. B 
sin. (S—4)= sin. § cos. 6—sin. 4 cos. 8 
Add these two equations, 
sin. (8+-4)+sin. (8 —é)=2 sin. 8 cos. 4 
Subtract sin. (8 — 4) from each member, 
sin. (8+6) =2 sin. 8 cos, 6~— sin. (8 — 6) 
Let 8=n 4, the above becomes 
sin. (2-+1)§=2 sin. 2 4 cos, é— sin. (n— 1) 4... (m) 
In the above formula, let n =1; .. n+1=2,2—1=0 
*, sin. 2 6=2 sin. 4 cos. 6 —sin. 0 
=2 sin. 6 cos. 4, the same result as in (g). 
Lét n=2 ; .0.n+1=3,n—1=1; 
sin. 3 6 =2 sin. 2 4 cos. 6— sin. 6 
=2X2 sin. 6 cos. 6X cos. 6 — sin. 4 
=4 sin. 6 cos.26—sin. 6 « 
=4 sin. é (1 —sin.’ 6)—sin. 4 
=3 sin. é—-4 sin.° 6 - - (x) 
Let 2==3;..°. xX 1=—4, n——1=2: 
*. By formula (m): 
sin. 4 6 =2Xsin. 3 6Xcos. 6— sin. 2 4 
=2 (3 sin. 6 —4 sin.* 4) cos. 6—2sin. 6 cos. 4 
=(8 cos.* 6 —4 cos. 4) sin. 4 


It is manifest that, by continuing the same process, we may 
find in succession, sin. 5 4, sin. 6 4, 


> . si -h&c, 


To determine the cosine of (n+1) 4, in terms of n 6, (n—1) 4, 
and 6, 


By formula (c) and (d) : 
cos, (8 +4)=cos. 8 cos. 6—sin. 8 sin. 4 
cos. (8 —4)=cos. Scos. é6+sin. f sin. 6 
Add these two equations, 
cos. (8+4)+cos. (6 — 6)=2 cos. 8 cos. 4 
Subtract cos. (8 — 4) from cach member, 


yn) ver? pe 


guns 


52 ANALYTICAL PLANE TRIGONOMETRY. 


cos. (8+4)=2 cos. 8 cos. 6 — cos. (8 — 4) 
Let S=n 4, the above becomes 
cos. (n+1) 6=2 cos. n 4 cos. 6— cos. (n—1)4- - (0) 
In the above formula, let n=1;. .. n+1=2, n—1=0; 
Then, cos. 2 6=2 cos. 6 cos. 6—cos 0 
=2 cos.” 6— 1, the same result as in (A2). 
Let 7n=2, .°. n+1=3, n—-I=I1; 
“. cos. 3 6=2 cos. 2 4 cos. 6— cos. 4 
=2 (2 cos.” 6 — 1) cos. 6— cos. 4 
=4 cos..6—3cos.6 - - - - (p) 
Let n=3; .°. n+1=—4, n—1=2; 
cos. 4 6=2 cos. 3 4 cos. 6— cos. 2 4 
=2 (4 cos.* 6 —8 cos. 4) cos. 6 — (2 cos.” 6 —1) 
=8 cos.‘ 6— 8 cos.? +1 

It is manifest that, by continuing the same process, we may 
find, in succession, cos. 5 4, cos. 6 8, cities LANES, 

By adding and subtracting (a) and (b), and by adding and 
subtracting (c) and (d), we obtain the following formulz, which 
are of considerable utility. 

sin. (6+)+sin. (6—f8)= 2 sin. 4 cos. B) 

sin. (6+8)—sin. (6—f§)= 2 sin. 8 cos. é 

cos. (6+8)+cos. (6—8)= 2cos.dcos.B( ~~ ~~ (7) 
cos. (6+8)—cos. (6—@)= —2 sin. 4 sin. 6 


Any angle 4 may, by a simple artifice, be put under the form, 


6+8 6—f 
(cae ee eS SRS 
2 + 2 
And, in like manner, 
asta h saps 
2 2 
r ee 
.. sin, 6=sin. a 
tai = 6 
yt EE os sae ti sin. B cos. = - - -(1) 
Min Oosaid ae a 
tO a Be | 
— s&s 6 
=sin. TF ros. =F —sin, ae _ - - - (2) 


cos. 6=cos. ‘pee aoe 
“648 —B . 648 . §—8 
4 


= COS. Tie ada — Sin. my sin. 


ANALYTICAL PLANE TRIGONOMETRY. 53 


cos. 8=cos. alge a ” | : 
=COB. Raley Rats + sin. es sin, - +. (4) 

Add together (1) and (2): 

sin. 6+sin. B=2 sin. TF cos. es - = = = © = (r) 

Subtract (2) from (1), 

sin. é— sin. S=2 sin. BE genet Sn oe en fa) 

Add together (3) and (4), 

cos. +cos. 8=2 cos. ae a ae $y. dntie- 8, spaade oft) 

Subtract (4) from (3), 

cos. 6—cos. B= —2 sin. oe sin. aa - : -/(2) 


These formule, which are of the greatest importance, might 
have been immediately deduced from the group (q), by chang- 


; ‘ ahh 
ing 6+8 into 4, 6—8 into 8, 4 into =, into 


2 
Divide (r) by (s) : 
2 sin. cae Cos. smile 
sin. 4+sin. 8 th 2 2 
sin. @—sin.B 4. °—8 6+8 
2 sin. 5) COS. 3 
tan. ke : i (+48) 
an 
See ee Ne et Se (2D) 
tan. Pa tan. 4 @—f) 


Multiply (a) by (2); then, 
_sin. (6+) sin. (8 —8)=sin.’ é cos.” 8 — sin.’ B cos.” 6 
=zhit, MAeetsih,® ite are tier Ske 
Multiply (c) by (d); then, 
cos. (6+) cos. (3 —8)=cos.? 4 cos.? 8 —sin.? 6 sin.’ 8 
=cos”2é@—sin? 6B - - - + (y) 


bee wae e 


aA ANALYTICAL PLANE’ TRIGGNOMETRY. 


We will now investigate a few properties where more than 
two arcs or angles are concerned, and which may be of use 
in. the subsequent part of this work. 


Let 4; 8, y,.be any three arcs or angles. 


Then, 
in. 6 sin. y-tsin. 6 sin. (6+8+y) 
RTE ia sin 7s B sin. (6+8+7) 
For by formula (a) 
sin. (6+-8+y)=sin. 4 cos. (8-+y)+cos. 4 sin. (0+y) which 

[putting cos. 8 cos. y—sin. f sin. y, fur cos. (8+y)], is=sin. 4 
cos. 8 cos. y-—sin. 4 sin, 8 sin. y+-cos. 4 sin. (8+y) ; and, mul- 
tiplying by sin. 8, and adding sin. 6 sin y, there results sin. 4 
sin. y+sin. 8 sin. (6+8+y)=sin. 6 cos. 8 cos. y sin. 8+ sin. 6 
sin. y cos.? 8+cos. 8 sin. 8 sin. (8+y7)=sin. 4 cos. 6 (sin. 8 
cos. y-+cos. 8 sin. y)+cos. 4 sin. 8 sin. (8+y)=(sin. 6 cos. 8 
+cos. é sin. 8) sin. (8+y)=sin. (6+) sin. (8+7). 

Hence, dividing by sin. (6+), we have, 

’ in. 6 sin. in. in.(4 
es eu aes sin 7+sin 6 sin.(6+8+y7) 
In a similar manner it may be shown, that 
__sin. 6 sin. y——sin. 8 sin. (6—8+7) 
7 sin. (6 —B) 

If 6, 2, y, 6, represents any four arcs or angles, then writing 

y+6 for y in the preceding investigation, there will result 
ae (6+. Hee é sin. (y+0)+sin. 8 sin. 6+8+y+0) 

; Yi sin. (6+ 8) 


A like process for five arcs or angles will give 
in. 4 sin. (y-+é in.Bsin.(8 a 
sin (8 eyo eye tee Sin are) Sin Pein Er Bt eke +2) 


sin. (6+/) 
And for any number 4, 8, y, &e. - - - % 
sn (Gee te..p) Sind sin. (y+9+...A) tsin.f sin.(@+8 ty +...) 
aD i. 


Taking again the three 4, 8, y, we have 
sin. (8 —y)=sin. 8 cos. y— sin. y cos. 6 
sin. (y— 46) =sin. y cos. 6—sin. 4 cos. y 
sin. (6 — 8)=sin. 4 cos, 8 —sin. 8 cos. 4 
Multiplying the first of these equations by sin. 4, second by 
sin. 8, third by sin. y; then adding together the equations 
thus transformed ; there will result, 
sin. 6 sin(S—y)-+sin. § sin. (vy — 4)-+sin. y sin. (6—8)=0 
sin. 6 sin. (8— y)-++cos. 6 sin. (y—4)+cos. y sin. (6--8)=0 
These two equations resulting from any three angles what- 
ever may evidently be applied. to the three angles of any 
triangles. | 


ANALYTICAL PLANE TRIGONOMETRY. 535 


Let the series‘of arcs or angles 4,8,y,56 - - - - - 2, 
be contemplated, then we have formula (x) 


sin. (6+) sin. (6—8)=sin.’ 6 — sin.’ 8 

sin. (8+y) sin. (8 —y)= sin.’ 8 — sin.’ Y- 

sin. (y+8) sin. (y — 4)=sin.” y — sin.’ 6 
Odds es =< ce. . 

sin. (A-+4) sin. ‘(4 — 4) sin. *}—sin.? 4 


Adding these equations together, we have 

sin. (2+) sin. (8 —8)-+sin. (P+y) sin. (B—y)+sin. (y+) 

Sin. (Y—O) +. -e ewe ees sin. (A-+4) sin. (A— 6)=0 

Proceeding in a similar manner with the sin. (6—§8), 
cos. ()+f), sin. (8—y), cos. (8+y7), &c., there will at length 
be obtained cos.(8+8) sin.( — 8)-+cos. (3-+7) sin. (S— ae Fe 
cos. (A+) sin. (A—4)=0 

If the arcs 4, 8, y- - - - % form an arithmetical progression 
of which the first term is 0 the ratio e and the last term A, any 
number n of circumferences, then will S—4=2, y —8=s, &c., 
6+f8=2, 8-+y=32, &c.; dividing the whole by the sin. ¢, the 
preceding equations will become 

sin. g-+sin. 3e+sin. 5¢+é&c.= 
cos. ¢-+cos. 3g-+cos. 5e-+&c. ai 
If were equal 22, these equations would become 
sin. e+sin. (e-+%)+sin. (¢+2%)+sin. (¢+32)+d&c.=0 
cos. e-+cos. (e+£)+cos. (e+ 2%)+cos. (g+3%)+&c.=0 

The last equations, however, only show the sums of the 
sines and cosines of arcs or angles in arithmetical progression 
when the common difference is to the first term in the ratio of 
2tol. To find a general expression for an infinite series of 
this kind, let 


S+sin. 6+sin. (6+) -+sin. (64+28)+sin. (6+38)+ - - - - &c. 
Then since this series is a recurring series whose scale of re- 
lations is 2 cos. S—1, it will arise from the oeve Duman of a 
fraction whose denominator 1—2y cos. 8+ x? making y=1 
Now this fraction will be, 
_sin, 4+ /sin.(6+8) — 2 sin. 4 cos, & 
1 — 2xcos. 8+ x? 
Therefore, when x=1, we have, 
g_ sin. 6+sin. (6+8)—#% sin. 4 cos. 8 
2—2 cos. 8 : 
And this because, 2 sin. 4 cos, 8=sin. (4+8) +sin. (§—f) 
__sin, é— sin. (8 —8) 
2 (1—cos. 8) 
Now putting 4 for (8+) and 8’ for (6--8) we have from 
formula (s) : 


by formula (q). 


56 ANALYTICAL PLANE TRIGONOMETRY. 


sin. 4’— sin. 8’=2 cos.1 (4’+8’) sin.2 (6/— 8’) 
Hence, it follows that, 
sin. 6—- sin. (6-~ 8) =2 cos. (6 — 18) sin.38 
Besides which we have, 
1 — cos. S=2 sin.718 
Consequently the preceding expression becomes, 
S=sin. 6+sin. 6+8)+sin. (6+28)+sin. (6+38) &c. ad 
infinitum, cos. (9 — 18) 
QsiniB - - - - - = = (22) 
To find the sum of n+1 terms of this series, we have simply 
to consider that the sum of the terms past the (n+1)th, 
that is the sum of | 
sin. (6+(n+1)8) +sin. (6+(n+2)8) +sin.(8+(n+8)8)+ 
&c. ad infinitum, is by the preceeding theorem, 
2008 Chit He) 
ba 2 sin.t 8 
Deducting this from the former expression, there will remain 
sin. 6+sin. (6+8)-+sin. (+28)+sin. (0+38)+ - - - - 
- - - - sin. (6+n8)=cos. (6—38)—cos. (6+(n+1)8) 
2 sin. ifs 
__ sin. (+3 28) sin.t (n+1)8 
iP sin. 18 (23) 
By like means it may be found, that the sum of the cosines 
of arcs or angles in arithmetical progression, is cos. 6+cos. 
(+8)+cos. (0+28)+cos. (6+38)+ &c. ad infinitum, 
_ _ sin.(8 +438) 


2 sin. 38 - - = (24) 
Also, 
cos. 6+cos. (+8)-+cos. (0+28)+cos. (4+388)+-- . - 
- - - (cos. +n) cos. (6+18(sin.2 (n+1)8 
cok panic, 12 ER SC, 


To find the numerical value of the sine, cosine, &c., of 45°. 


In the circle ABD, draw CA, CB, radii at B 
right angles ; join AB. 
_ Then by Definition (12) 


1 AB A D 
Chord ABC (90 ey 
sis | AEB? 
Chord’ 90 = AGH E 
AC?-+- BC? _ 


weAC 


‘” 


“ANALYTICAL PLANE TRIGONOMETRY, 57 


PEND ooo, 
‘ ca AC? > CSAC 
a PE an eect ve @) 
Now, the chord of an arc is equal te twice the sine of half 
the arc; therefore, 

2 sin. 45°=chord 90° 
4 sin.” 45°=chord? 90° 

=2, by Equation (1) ; 


“ve sin. a Vs ” 


J 2 
Again, by. table I. : 
sin.? 6+cos.? 6=1 
cos.? 45°—=1 — sin.” 45° 
1 


ay Gata) 


1 


—- —— 


2 
cos. 45°= oye sin. 45°. 


Also, 
in. 45° 
tan. 459 ee mn 
cos. 45° 


=]=cot. 457. 


To find the numerical value of the sine, cosine, &:c., of 30°. 


In the circle ABD, draw CP, making with Pies 
CA the angle ACP=60° ; join A, | 
Now, 


2 sin. 30°=chord 60° 2 " 
py3s 
AC z 
soe .- AP=AC, + the triangle APC 
is equiangular, and therefore 
equilateral. 
oe 
sin. spose. 
z 
Again, 
cos. 80°= /] — sin,? 30° 
=V] iene bs V3 
2 2 


58 ANALYTICAL PLANE TRIGONOMETRY. 
Also, 


sin. 30° 
tan. 30°= ew 
cos. 30° 
— 1 e 
oo. | 
] 
O07 = ae ee 
‘ tan. 30° wae 


To find the numerical value of the sine, cosine, &c., of 60°. 


sin. 60°=cos. (90°— 60°) 
—conrad., 


= ae by last art. 
2 


Again, 
cos. 60°==sin. (90°— 60°) 
=sin. 30° 


Also, 
tan. 60°= V3 
} 
cot. 60°= V3 


Jt is required to find the sum of all the natural sines to every 
minute in the quadrant, radius =1. In this problem, the actual 
addition of all the terms would be a very tiresome labor, but 
the solution by means of formula (z8), is rendered very easy. 


sin. 1( w+ 1)S=sin. 45°, 0’, 30” and sin. To sifi. 30", 
sin. 45° sin. 45° 0’ 30” 


————————————— 


sin. 39” 


Applying that formula we have sin. (6+34n€)=sin. 45°, 


=3438.2467465, the same sum required. 


Let it be required to find the sum of the sines to every mi- 
nute of the are of 60°. 


Here the numerical expression in the equation would become 
30° 30° 0/30" 
Sip. oie sine FOR" = 6.5001 Bee Rdvo1Ne4 5958 


sin. 30” 
= 1719.123373.25 equal the sum of all 
the natural sines to every minute of the arc of 60°. 


It may be useful to exhibit the most useful results in this 
chapter, in the following table. 


ae 
ANALYTICAL PLANE TRIGONOMETRY, 59 


TABLE III. 


(1.) sin. @+8) =sin. 6 cos. Psin. B cos. 4 
(2.) cos. (8) =cos. dcos. P#sin. 4 sin. 8 


~/-etan..6—tan. 
(3.) tan. (648) ones 


1=¢tan. é tan. 8 


(4.) sin. 2 4 =o win © Gos a 
(5.) cos. 2 =cos.7d—sin.28=2 cos.? 6—1=1I—2 sin.” 9 
2 tan. 4 
6.) tan. 2 4 —=———_ — 
©) 1— tan.’ 4 
: 6b L—— cos: 4 
de = Nae : 
(7.) sin 5 5 
(8.) cos. a — a / Leos: 8 
2 2 
(9.) tan if ER A Lae. 6 1—cos.6 sind 
2 1+cos.6 sin.  1+cos.d 
6 6 
10.) sin. 4 = i a 
(10.) si 2 sin g C8 5 
(11.) sin. 3 4 =3 sin. 6—4 sin.* 4 
(12.) cos. 3 4 =4 cos.* —3 cos. 4 


(13.) sin. (n+1) 6 =2 sin. né cos. 6 —sin. (n—1) 9 
(14.) cos. (n+1) 4 =2 cos. nd cos. 6—cos. (n—1) 9 


: ; eee § — 
(15.) sin. 6+sin. 8 =2 sin. hs COR os 
, . b— 8-++f 
(16.) sin. é—sin. 8=2 sin. COs. tak 
y (ale S| 
(17.) cos.é+cos. S=2 cos. cos. —5 
Piety . b— 
(18) cos. é6—cos. 8= — 2 sin. sin = 
(19.) sin. é+sin. 8 “" 2 
sith 6-—sini 8 §—p 
tan. 2 


(20.) sin. (6+)+sin. (6 —§)=2 sin. 4 cos. B 
(21.) sin. (6+8)—sin. (—§)=2 sin. 8 cos. 4 
(22.) cos. (6+)-+cos. (46 — 8)=2 cos. 4 cos. B 


tug 


."Y 
ee 


a 4 
60 ANALYTICAL PLANE TRIGONOMETRY. 


(23.) cos. (6+()—cos. (4 —)= — 2 sin. 4 sin. 8 
(24.) sin. (6+ 8) cos. (6—)=sin.’ é—sin.? B=cos.? B— cos.? 6 
(25.) cos. (8+) cos. (6—8)=cos.? 6—sin.?8 =cos.?4+cos.28—¥ 
(26.) sin. 6+-sin. (0+8)+sin. (6+28)+sin. (6+388)+ ---- - 
Ween (be) = sin. (6-+27f8) sin. 1 (w+1)8° 


sin. 1 8 
(27.) cos. 6+cos. (6+8)-+cos. (6+28)+cos. (64+88)+ - - - 
- - - cos. (+78) _cos. (6+38) sin. 3 (n+1) B 
a sin. 2 8 
: | 1 
(28.) sin. 45° =COS. cab eg 
(29.) tan. 45° =cot. 45°=1 
: 1 
(30.) sin. 30° ==COS. 60° =—- 
3 
(31.) cos. 30° =COSs. 60°= 
1 
o ee 5 eee See 
(32.) van. 30 =cot. 60 73 
(33.) cot. 80° =tan. 60°= V3 


The formule of Trigonometry may be multiplied to almost 
any extent, and tl same quantity may be expressed in a vast 
number of different ways. An intimate acquaintance with 
those given in the above table is essential to the progress of 
the student. 


The following, although of less frequent occurrence, may 
occasionally be found useful, and ean be readily deduced from 
the above. 


(34 sin. (45°-E6) cos. =Esin. 6 
Dy cbs. ee a) a eae 


eS 
(35.) tan. (45°10) _< itetans? 
1=tan. é 

8 1+sin. 6 
36.) tan.? jt = a. 8 
Cages). ~ sages 

6 1+sin. 6 oo 
37.) tan. ode = ‘id ; 
(37.) tan. (4504. 5) See “ee 
(38.) sin. (6+) tan, é-+ tan. 8 cot. S=ecot. 6 


sin. (6—() ~ tan.é—tan. 8 cot. B—cot. 6~ 


_— or "a neil 


: a i 
at a 


@. ro 
yt . 
ANALYTICAL PLANE TRIGONOMETRY. 61 
' ae (0+-8) cot. B—tan. 4 cot. é—tan. 8 
” Cos. (6—) = cot. B-+tan. § cot. sot. d-+tan. B 
(40.) Sin. @+sin. 8 tan, SFP 
cos. 4+cos. 8 2 
(4y,) Sine ttsin. BL oot SP 
cot. d—cos. . 2 
(49, sin tosin. Bs tan, SB 
cos. 6-+cos. 8 2 
(43. ) a ae = —cot. +8 
cos. 6— cos. 8 2 
(44.) se met 2 =- + cot. secre 2 
cos. §—cos. 8 2 2 
(45.) tan. 6+ tan. 8 sin. (8 + °) 
cos. 4 cos. 8 
(46.) cot. 6+ cot. B = sina @ +8) 
sin. 4 sin. 8 
(47.) tan. — tan. out: 8) 
, cos. 6 cos. 8 
(48.) cot. 6—cot. 6 se! ype ioe od, 
sin. 6 sin. 8 
(49.) tan” 6—tan* @ = sin. (448) sin. (6@—8) 
cos.? 4 cos.’ 8 
(50.) cot.” 6—cot.’ 8 Paya eG +8) sin. (° ash 
fe sin.” 6 sin? 8 


In order to become familiar with the various combinations, 
and dexterous in the application of these expressions, the stu- 
dent will do well to exercise himself by verifying the follow- 
ing values of Sin. 4, Cos 6, Tan. 4, which are extracted from 
the large work of Cagnoli. 


6* 


62 — 
tt 


v2 


1. cos. 6 tan. 4 | 
16. 
9 cos. 4 
’ cot. 4 17 
3. /1—cos.’ 6 18, 
a 
4. /1+cot.? 4 19. 
; . 6 
a 20 
V1-+tan.? é : 
Pea oe 
. 2sin. 908-5 91. 
"7. Vv 1— cos. 24 ao 
2 
ue t 6 
pa ST. 
1+tan.’ ; 
an. 5 94 
2 
9. 6 F 
cot. 5 +tan. 5) sek 
10, 22: (30° +4)—(sin. 30°——#) 
: WE 
in.*(45° + : )\—1 
11.22:sin: 5 si 
A 
12. 1—2 sin.’ (45°— 5) 
: 7 
1—tan.’ (45°— >) 27. 


13. 


14. 


15. 


‘3 


ANALYTICAL PLANE TRIGONOMETRY. 


TABLE OF THE MOST USEFUL ANALYTICAL VALUES OF 
SIN. 6, COS. 6, TAN. 6. 


VALUES OF SIN. 4. | 


6 
1+tan.’ (45°— 5) 


28. 


Y 
oO a ase i ACh mae 
tan.(45 +5) tan.(45 S 


8 é 
tan.(45° +5) + tan.(45°—5) 29. 
sin. (60°+0)—sin. (60°—#) | 30. 


VALUES OF cos. 4, 
sin. 4 
tan. 4 


. sin. 4 cot. 6 


v¥1—sin.? 4 
cate: ae 
Vi1-+tan.’ 6 
cot. 6 
Vi+cot.d 
8 b 


cos.’ ~—=sin.*— 
2 2 


1—2s) ae 
ory 


6 
. 20s." wen 


1—tan.? 


1+tan. 5 oe: 


cot. Y —~fan. 


6 
cot. 5) + tan. 9 
1 
eee et 
1 + tan. 6 tan. — 
2 
2 


——_—____, 
tan.(45°-+5)+cot. (45°+5) 


6 ) 
2 cos.(45° +5)cos.(45°— 5) 


Sa 
cos. (60°+4) +cos. (60°—8) 


i tie 


. - “ 
We . 
%, » 
~~ 


ANALYTICAL PLANE TRIGONOMETRY. 63 
4A, 


VALUES OF TAN. 4, 


sin. 4 
31. cos. 4 2 cot. oY . 
ote j 
1 cot.2- —1 
._-—-- a 
32 cot. 4 2 
ry Dia on Rt et 
cot. — — tan. = 
33. cos.” he 2 2 
39. cot. 6—2 cot. 2 
sin. 4 1—cos. 24 
84. [aa ee ee | aE. Waid ais 
Vv 1—sin.? 6 of: sin. 24 
sin. 24 
35. ee a: 1+cos. 2 4 
pee St cos. 26 
ah b 42. l-+cos. 246 
an. 3 ‘ 4 Ms s 
Rotate At 43. tan.(45 +5)— tan. (45 ry, 
|— tan.’ 2 eA Le 4h ee 


From certain properties of the circles to be discussed in an- 
other volume, other important trigonometrical formule, may be 
deduced, furnishing us with more expeditious means of deter- 
mining, numerically, the values of some of the trigonometri- 


cal lines, and ratios, all of which will occur in their order. 


To develop sin. x and cos. x ina series ascending by the 
powers of x. 
The series for sin. z must vanish when z=0, and therefore 


no term in the series can be independent of z, nor can the 
even powers of x occur in the series; for if we suppose 


sin. t= a,t+a,0°+a,0°+a,¢'+a,0°+ 2... 
} = 2 8 4 5 
then sin. (— x) =—a,x+a,2°— a,2°+a,0'—a,0+ ... 
but sin. (—2)=— sin. « 
as 2 8 4 5 
oe as, aot Cas Og mm ig seus 5 hence a,—0, a,=0.... 
“SIM. t= @,0-+a,0'+a,2°+a,0°+ - - - «= - (1) 


* 


64 ANALYTICAL PLANE TRIGONOMETRY. 


Again, the series for cos. x must = 1 when 2 = 0, and there- 
fore the series must contain a term independent of 2, and it 
must be 1; also the series can contain no odd powers of a, for 
if we suppose 

608. @— lato. tba ta,z-t 2. es 
then cos. (—r)=1— a,%#+a,2°—a,a*+a,v'— .... 

but cos. (—2)=cos. x 


== 1a aba t+O,t +-¢,0 +" Sr. ae 
@,=—— 4,, dj=—d5,...  .*. dy=0, 2520 3. 
. Cos t—=)-Paxtt+a, a +aje+ - - -- - = (Q) 
Hence cos. ¢+sin. =1+a,2+a,0°+a,°+a,0'+a,2° - - (3) 
cos. & — sin. t=1— a,t&+a,2°—a,2°+a,c'—a,a°+ - - (4) 


Now in equation (3) write « +h for x, and we have 
cos.(e+h)+sin(x+h)=1+a, (e«+h) +a, (e+th)’+a,(«+h)'+ (5) 
but cos. (e«+h)+sin. (w+h)=cos. x cos. h—sin. x sin. h 
+sin. x cos. h+cos. x sin. h 
=cos. h (cos. +sin. x)+sin. h (cos. e—sin. 2) 
rel Lt-c, to hit. }).(1 4-0 t+, 2° +40 ey ) 
+. (a,h+a,h’+a,h'+.. ee ve estat Diva) 
Fa ede +a ek 4+, 
| +a,h—a,*th+a,a,vh+ ... 
= TR Maat ECP ee - - - (6) 
| +a,h’ + 
a ett -aisal 
Comparing the tee (5) and (6) we have " 
Ita,e+a,2" +a,2° Te l+a,t+a,27 +a,7 + 


+a, h+2a,ch+38a,0°h+ +a,h—a,a,th+a,a,x°h— 
+ a,’ +3a, th’-+- -= +a, h? +a,a,ch'+ 
= a,h’ an ‘e + a,h* car | 


+J a J 
and equating the coefficients of the terms involving the same 
powers of z and h, we have 


———q,a.; therefore a, =——— = — 
2a, 1443 2 2 1.2 
a,a, a,* 

— a a ° s ° e a = a EN ha 
3a 12 3 3 1.2.3 


Bag cee Sogo Gan Akg eed 


+ 
- 


ANALYTICAL PLANE TRIGONOMETRY. 65 
Fie : ay” e ag 
hence sin, e=a,2 er 23° t T2345? 71.2.3.4.5.6.7% * 
r a,* 2 a 4 ay 
cos. t= geo 2 i 9347—19.3.45.6° 7 
To effeet 


and we have only to determine the value of a,. 

this, we have 
in. =a, z $y, ae Bucs aif 
ea ede he Leg, ae 

Ae e e ° 
=a,2(1— mec 3g rp 

Now the value of z may be assumed so small that the se- 
ries in the parenthesis, and sin. 2, shall differ from 1 and z re- 
spectively, by less than any assignable quantities ; hence ulti- 


mately 
x=a,z, and therefore a,=1; whence 
: i 7a bok 
SIMs ho og. TaaAb TBAB GT me te 
2. a" x 
Ey os 


pon th gt hao ea oe Tae 


To develop tan. x and cot. x in a series ascending by the 
powers of X. 
The development may be obtained from those of sin. x and 


cos. z, already found. 


ps 4b & 
nm 2 t— 159 10.5 99 23 

tan. r= = 2 x 
oes &e 


cos. x : Las 
DATUypes ee ae 


and the series will therefore be of the form 
PR ee oe 


a , + ~ 
i pod 
he 1.2.3.4.5 
Hence, let z+a,27°+a,2°+..= xr 
Prebrassnn Mir ss Sai 


A oo 
‘ ° 183 TT 284.5 
= (1—- -z3t Ory an »-) (ear +a,2'+. ) 


66 ANALYTICAL PLANE TRIGONOMETRY. 
Sago ets ® ae 4 La 


1 
po es Be 
+ 12.34 


Hence, equating the coefficients of the like terms, we have 


go Pain 1 4s pcx 
sel Sy OE pos ts! 59.9 
Oe 1 1 2° 
a, = Te ke EE oa a ee = lips “ 
T1979 345 loads “ts = pega e 
. 22° OT a 
tan a+ 1.2.3 -|- 1.2.3.4.5 * eS Mie, a eT Welt e 
a x oy* 


* apenas, 2 
oO Se eee 1O5e a Sas 


CHAPTER III. 


FORMULZ FOR THE SOLUTION OF TRIANGLES. 


We shall here repeat the enunciations of the two proposi- 
tions established in Chapter I. 


PROPOSITION I. 


In any right-angled plane triangle, 

Ie. The ratio hich the side opposite to one of the acute 
angles has to the hypothenuse, is the sine of that angle. 

2°. The ratio which the side adjacent to one of the acute 
angles has to the hypothenuse, ts the cosine of that angle. 

3°. The ratio which the side opposite to one of the acute 
angles has to the side adjacent to that angle, is the tangent of 
that angle. 


Thus, in any right-angled triangle ABC, 


« 

a 

. Fae 
2 


ANALYTICAL PLANE TRIGONOMETRY, 67 


Cc 
ee = sin. A, AG = 098 f.® —_ tan. A 
=> cos. C, =) Siow, = 60t..C 
Or, CB= ACsin. A) ) 
= C cos. | | 
BA = AC cos.A) ! B A 
SAGO kin’ ok ws isla mak a OR a? 
CB= BA tan. A 
= BA cot. C 


PROPOSITION II. 


In any plane triangle, the sides are to each other as the 
sines of the angles opposite to them. 


We shall, frequently in treating of triangles, make use of 
the following notation ; denoting the angles of the triangle 
by the large letters at the angular points, and the sides 
of the triangle opposite to these angles, by the corrcs- 
ponding small letters. | 


Thus, in the triangle ABC, we shall 
denote the angles, BAC, CBA, BCA, G 
by the letters, A, B,C, respectively, , 
and the sides BC, AC, AB, by the 
letters a, b, c, respectively. 7 
According to this, we shall have, by 
the proposition, 


Loa sin. A) 
bas.” sina B 
a sin. A 
ve gine Ra aa whitgs TG 
b x sin. B 
o- 1 sips CO 


ae : 
68 ANALYTICAL PLANE TRIGONOMETRY. 


PROPOSITION IH. 


In any plane triangle, the sum of any two sides, is to their 
difference, as the tangent of half the sum of the angles opposite 
to them, is to the tangent of half their difference. 


Let ABC be any plane triangle, then, by 


Proposition {I £ 
Mal sin. A 
6b” sin. B 
a+b sin. A+sin. B A B 


———— 


(ae A—sin..B 


But, by Trigonometry, Chap. Il. (r) 


in Ath Era A—B 
gin. A+sin. b= 2 sin. 5) COS. 5) 
: oy A——ts 
sin. A—sin. B = 2 cos. sin. —>— 
2 sj A+B <A—B 
eae RARE, PETIT. 
ab A+B . A—B 
cos. 5 — sin. —5 
A+B A—B 
= tan cot. 
2 
At 
tan +Bi 
bi! 2 
(Noes a 
tan. 5) 
And in like manner, mae 
di-c tan ee 
(ERENT ENE § 0 IS 0 Sy AN, 
tan. 5) 
; B+C : 
b+ an.—3 
ro 
B—G 
tan 


¥* 
ANALYTICAL PLANE TRIGONOMETRY. 69 


PROPOSITION IV. 


To express the cosine of an angle of a plane triangle in 
terms of the sides of the triangle. 


Let ABC be a triangle ; A, B, C, the three c 
angles ; a, b, c, the corresponding sides. 
1. Let the proposed (A) be acute. 
From C draw CD perpendicular to AB, the 
base of the triangle. 
Then, AL: 
BC?=AC?+AB’—2AB. AD (Prop. XX VI. 
B. IV. El. Geom.) 
Or, 
a= +c ~—2c.' AD 


But, since CDA is a right-angled triangle, 
AD = AC cos. CAD = b cos. A 
a= BW + c’*—2be cos. A 
P+ e—a 
2be 
which is the expression required. 


. cos. A = 


2 Let the proposed angle (A) be abtuse. 
From C, draw CD perpendicular to AB producea. 
Then, 
BC *= AC’ + AB® + 2AB.AD 
Or, 
a? == bP ier Be AD 
But, since CDA is aright angled triangle, pD~A 
AD = AC cos. CAD 


= AC xX — cos. CAB 
-.° CAB is the supplement of CAD. 


=—bcosA 
Oe =e — 25608. 
; Ais b+ ce a’ 
...cos. A = She 


It will be seen that this result is identical with that which 


we deduced in the last case, so that, whether A be acute or 
Obtuse, we shall have, 


7 


* 


ag 


70 ANALYTICAL PLANE TRIGONOMETRY. 
B+ ce? — - a > 
cos. A = “ae 


Proceeding in the same man- 
ner for the other angles, we shall 


find, sebfoab nates: 
oo OF 
gs ie 
ac 
e+h—ec 
cos. O =. “basin ame 


PROPOSITION V. 


To express the sine of an angle of a plane triangle in terms 
of the sides of the triangle. 


Let A be the proposed angle ; then by last prop., 


Y+ce—@a@ 
cos... A = spent 
Adding unity to each member of the equation, 
A= ieee ee 
1 + cos. A = ari 
ee b?+2bce+c’—a’? 
x Bee 
Orb cy aE 
aa 2Q2bhe 
(b+c+a) (b+c—-a) ; 
Pe age pce Pot. aaa rt ee 
; b? +c? — az 
Again, cos. A = Te pes 
Substracting each member of the equation from unity, 
te b?+-c’—a? 
? 1—cos. A = 1— bie 
Qbc +o 
x 2he 
_ &—(b’°—2be+c*) 
2be 
fd Shes 
or 2Qhe 


(a+b—c) (a+c—b 
zilashdr th (ck onnd) gos i eeosbaluis 


ANALYTICAL PLANE TRIGONOMETRY, 71 
Multiplying together equations (1) and (2,)~ 
(1+c08A).(1—~cosA) = +) OF) ape b) (abe) 
But (1+cos. A) (l—cos. A)= 1—cos.’? A rs 
= sin.” Av #(Table 1.) 
wh _(atb+e) (b+c—a) (frrob) (a+b—c) 
4b’¢ 
Extracting the root on both sides, 
1 


sin. A= Qhe / (a+b+c)(b+c—a) (a+c—b)(a+b—c)..(3 


The above expresison, for the sine ofan angle of a triangle 


in terms of the sides, is sometimes exhibited under a form 
somewhat different. 


Let s denote the semiperimeter, that is to say, half the sum 
of the sides of the triangle ; then 


at+b+c 


rai WK and, 2s = atb+e 

s—a = ee . 2(s— a) = b+c—a 

ti Gs ne Ha bd) 2c atec—d 

s—c = —— . 2 (s —c) = at+b—c 
Substituting 2 s,2 (s—a),...... naan ei b-+-c—a,..... in the 


expression for sin.’ A, it becomes 

16 s (s—a) (s—b) (s—c 
int A = 168 G=@) (82) (s=¢) 
4 b’c 


And extracting the root on both sides, 
ey ee ed ate 

BT tee DeS Mas (s—a) (s——b) (s—c) 

Proceeding in the same manner for the other | 


angles, we shall find E p> (8) 


sin. B= g's (s—a) (8—b) (S— | 


ae 
sin. = ab’ V's (s—a) (s——b) (s—c) 


72 ANALYTICAL PLANE TRIGONOMETRY. 


By equation (1) we have 
iS 7 1+ cos. _ = (at+b+c) (b+c—a) 
2be 
_ 48 (s—a) 
2be 


But, by Chap. II, 


; A 
1+cos. A=2 cos. 
A 4s(s—a) 
2 _ —— 
2 COS. 3 2 be 


Extracting the root on boti sides, 


ref oe oe re 


And in like manner, | 


B ee 

cos. —= s(s—b) eee aN 2 ie oh 
2 ac | 

ae ae 5 (s —s)-| 
2 ab ) 


By equation (2) we have 
1—cos. Aa late—b) (a+b —c) 
2bc 
_4(s—b) (-—0) 
ti 2bc 


But, by Chap. II, 
I1—cos. A=2 sin? 


: .,A 4(s—d) (s—c) 
gs ARN. ge 


Extracting the root on both sides, 


sin =A4/ ee 
? Cc 


And in like manner, 


sin — \/6=9 69 - 


sin =A/ 9D) 


io 
a? 


« 


- we have | 


~ 
e a 


* 


_ ANALYTICAL PLANE TRIGONOMETRY. —_73 


Dividing the formule marked (%) by those marked (c) 


tone = 4/ OS b) (s—9) 


e—9 |. 


tan = af, Seo) (0) ant ae >) 


s(s—b) 


, tan<-= J (s— 4 (s—?) | 


$s (s—c) 


RAE THR IV. 


ON THE CONSTRUCTION OF TRIGONOMETRICAL TABLES. 


Berore proceeding to apply the formule deduced in the last 
chapter to the solution of triangles, we shall make a few re- 
marks upon the construction of those tables, by means of which 


we are enabled to reduce our trigonometrical calculations to 


numerical results. 

It is manifest, from definitions 1°, 2°, 3°, &c. that the various 
trigonometrical quantities, the sine, the cosine, the tangent, &c. 
are abstract numbers representing the comparative length of 
certain lines. We have already obtained the numerical value 
of these quantities in a few particular cases, and we shall now 
show how the numbers, corresponding to angles of every de- 
gree of magnitude, may be obtained by the application of the 
most simple principles. , | 

The numbers corresponding to the sine, cosine, &c. of all 
angles from 1” up to 90°, when arranged in a table, form what 
is called the Trigonometrical canon. 

The first operation to be performed is * 


To compute the numerical value of the sine and cosine of 1’. 
» 


We have seen, Chap. II. formula (j) that 


b ws 
Se oll 1 — cos. 4 


5 


zo2aV1—sin. 6 


By which formula the sine of any angle is given in terms of 
the sine of twice that angle. 


7* 


a, 


* 


~ 


* 


g® 
44 


74 ANALYTICAL PLANE TRIGONOMETRY. 


6 i 
Now substitute PY for 4 and it becomes 
oy 6 ‘ be a/ b te 8 
in. 7 or sin. 53= 5 ta sat 3 
; ne / ; é 
In like’ manner, sin.5a— WE jet 1 1—sin. a 
att ny 4 W 
pi ra ye i / 1 — sin.’ 5F 
&c. = &c. 


ig 8 a/ ad 6 
And generally, sin. gat 4—-FV/ 1—sin.? re 
6 


Now let 6=30° Wg = 15° 


and applying the above formula, we have 
sin, 15°= a/3 —2 </,1——Sin se 0% 
But by Chap. If. sin. 30°=4  .*. sin.” 30°=?7 
° } ,° a aay WD A ee SS 
-. sin. 15 FY Seether i 
=3V2— v3 
=.2588190 - - - - 
Similarly, sin. '7°80'=/ 117] — gin? 15° 


= /1—1,/1— (.2588190) 
EST oe 
OCC. =40c6; 
It is manifest, that, by continuing the process, we shall ob-. 
tain in succession the sines of 3°45’, of 1952/30”, &c. 


In this way we find 
oO 


30 : 
Sin. ou OF sin. 1’ 45” 28” 7!” 30% =.0005113269, &c. 


3 


me, : ’ 
Sin. ou OF sin. 52” 44'” 3iV 45% —,0002556634, &c. 


From which it appears, that, when the operation above 
mentioned has been repeated so many times, the sine of the 


arc is halved at the same time that the arc itself is bisected: 
that is, 


*~ 
af 


ANALYTICAL PLANE TRIGONOMETRY. % 


The sines of very smail arcs are nearly proportional to the 
) arcs themselves. 


Hence we shall have | 
Sin. 52/44’ Ziv 45% 3 sin. I’ :: 52” 44’ giv 45v : y! 


wert 60 
ae ‘60 X 60 
:: 3600 : 4096 
Sgn racSite 52” 44" BY 45% X 4096 
, sre SiN. = 3600 
; __ 0002556634 x 4096 
ae 3600 
=.000290888204 ..... =cos. 89° 59! 


*.* sin. 6=cos. (90°—6) 
Again, *.. cos. = V1—sin. 6) 
cos. 1’= 71 —(.000290888204 ... )? 
= .999999915384 


The sine and cosine of 1’ being thus determined, we shall 
proceed to show in what manner we shall now be enabled to 
compute the sines and cosines of all superior angles. 

By formula (m) Chap. II. 

Sin. (n+1) 6=2 cos. 4 sin. 7 6— sin. (x — 1) 4 
If we suppose 6=1’ and z to be taken = to the numbers 1, 


era ths weve in succession, we find 
Sin. 2’/=2 cos. 1’ sin. 1’ — sin. 0=.0005817764 ... =cos. 89° 58’ 
Sin. 3’=2 cos. 1’ sin. 2’— sin. 1/= .0008726645... =cos. 89° 57’ 


Sin. 4’=2 cos. 1’ sin. 3’— sin. 2’/=.0011635526 ... =cos. 89° 56’ 
&c:= Ke. 
Again, by employing formula (0), Chap. II. 
Cos. (n+1) 6=2 cos. 4 cos. n 6—cos. (n— 1) 4 

If, as before, we suppose 6=1’ and n=1, 2, 3, ..... in succession, 
os. = 2 cos.’ 1’— cos. 0=.999999830 .... =sin. 89° 58/ 
Cos. 3’=2 cos. 1’ cos. 2’/—cos. 1’=.999999619 .... =sin. 89° 57’ 
Cos. 4/=2 cos. 1’ cos. 8’— cos. 2/=.999999328 .... =sin. 89° 56’ 

&c.=&e. 


It is manifest, that, by continuing the above processes, we 
shall obtain the numerical values of the sines and cosines of all 
angles from 1’ up to 90°. These being determined, the tan- 
gents, cotangents, &c. may be calculated by means of the re- 
lations established in table I. 

The above operations are exceedingly laborious, but require 
a knowledge of the fundamental rules of arithmetic alone. It 
is manifest that, in employing this method, an error committed 
in the sine or cosine of an inferior arc, will entail errors on the 


76 ANALYTICAL PLANE TRIGONOMETRY. 


sines or cosines of all succeeding arcs. Hence is created the 
necessity of some check on the computist, and of some inde- 
pendent mode of examining the accuracy of the computation. 
For this purpose, formule, derived immediately from esta- 
blished properties, are employed ; if the numerical results from 
these formule: agree with the results obtained by a regular 
process of computation, then it is almost a certain conclusion 
that the latter process has been rightly conducted. 
Formule employed for this purpose are called formule of 
verification, and of these any number may be obtained ; it will 
be sufficient for our present purpose to give one. 


Sine de cos? A=) <p Ae eansotig-= = Sere. tab. L 
And 2 sin. 6 cos. 6=sin. 2 4 
Hence sin. d=1V1+sin. 26+1V1—sin. 2 6 


cos. 6=1V1+sin. 26>1V1—sin. 2 4 

Now if we suppose =12° 380! 

sin. 12° 30’/=1V1+sin. 25° +4V1— sin. 25° 
cos. 12° 380’=1V1-+sin. 25° = V1— sin. 25° 

Hence, if the values of the sine and cosine of 12° 30’, and 
of the sine of 25° obtained by the method already explained, 
when substituted in these equations, render the two members 
identical, we conclude that our operations are correct. 

The values of the sine and cosine of 30°, 45°, 60°, &c. 
which were obtained in Chap. II, may be employed as for- 
mule of verification. 

We can obtain finite expressions, although under an incom- 
mensurable form, for the sines of arcs of 3°, and all the mul- 
tiples of 3°, 2. e. for 

Br, G2, 9°, 19? PS" 1G, 2h 2e , lag GU ce tied OU Brel os ie» 
45°, 48°, 51°, 54°, 57°, 60°, 63°, 66°, 69°, 72°, 75°, 78°, 81°, 
84°, 87°, 90°. 7 

We first obtain the values of the sines 30°, 45°, 60°, 18°, and 
from these we obtain all the others, by means of the formule, 
for 

Sin. (6+/), sin. (6— £8), &c. 

The numerical value of the trigonometrical functions have 
been calculated by some to ten places of figures, by others as 
far as twelve. We must have tables calculated to ten places 
to have the seconds and tenths of a second with precision, 
when we make use of the sines of angles which differ but 
little from 90°, or of the cosines of angles of a few seconds 
only. Tables in general, however, are calculated as far as 
seven places only, and these give results sufficiently accurate 
ior all ordinary purposes, 


ANALYTICAL PLANE TRIGONOMETRY. 77 


Such is the formation of the trigonometrical canon. The 
labor of the application of this canon, may be much facilita- 
ted by the application of a system of artificial numbers called 
logarithms, a description of which, forms the subject of the 
next chapter. 


CHAPTER V. 
LOGARITHMS. 
DEFINITIONS AND ILLUSTRATIONS. 


1. Logarithms are certain functions of natural numbers, by 
the use of which the tedious operations of multiplication and 
division are performed by the addition and subtraction of those 
functions; which consist of artificial numbers having such 
relations to certain natural numbers, that the sum of any two 
of those artificial numbers will be a similar function of the 
product of the natural numbers to which they have such rela- 
tion; and the difference of any two will be a similar func- 
tion of the quotient arising from the division of such natural 
numbers. 

Or more definitely, logarithms are the numerical exponents 
of ratios, being a series of numbers in arithmetical progression 
corresponding to another series in geometrical progression. 

Thus Me I, 2, 3,4, 5, 6, Indices or logarithms, 
?( 1, 2, 4, 8, 16, 32, 64, Geometrical progression. 

Or 0, 1, 2, 3, 4, 5, 6, Logarithms, 

; 3, 9, 27, 81, 243, 729, Geometric progression. 

Or OS yet2, so; 4, 5 Logarithms, 

, 1, 10, 100, 1000, 10000, 100000 Geom. progress. 

Where it is evident that the same indices answer for any 
geometric series, and therefore there may be an endless variety 
of systems of logarithms to the same natural numbers by 
changing the second term 2, 3 or 10, &c., of geometrical se- 
ries of whole numbers ; and by interpolation the whole system 
of numbers may be made to enter the geometric series and re- 
ceive their proportional logarithms, whether intregers or 
decimals. 

It also appears from the construction: of these series, that if 
any two indices be added together, their sum will be the in- 
dex of that number which is equal to the product of the two 
terms in the geometric series to which those indices belong. 
Thus, the indices 2 and 3 being added together, make 5, and 
the product of 4 and 8, being the terms corresponding to those 
indices is 82, which is the number corresponding to the index 5. 


b ee 


78 ANALYTICAL PLANE TRIGONOMETRY. 


In like manner, if any index be subtracted from another, the 
difference will be the index of that number which is equal to 
the quotient of the two terms to which those indices belong. 
Thus the index 6 — the index 4 is=2, and the terms corres- 
ponding to those indices are 64 and 16, whose quotient is=4, 
which is the number answering to the index 2. 

For the same reason, if the logarithm of any number be 
multiplied by the index of its: power, the product will be equal 
to the logarithm of that power. Thus, the index or logarithm 
of 4, in the above series is 2; and if this number be multiplied 
by 3, the product will be=6, which is the logarithm of 64, or 
the third power of 4. 

And, if the logarithms of any number be divided by the in- 
dex of its root, the quotient will be equal to the logarithm of 
that root. . Thus, the index or logarithm of 64 is 6, and if this 
number be divided by 2, the quotient will be 3, which is the 
logarithm of 8, or the equal root of 64. 

The logarithms most convenient for practice, are such as 
are adapted to a series increasing in a tenfold ratio, as in the 
last of the above forms, and are those which are usually found 
in most of the common tables on the subject. 


2. In a system of logarithms all numbers are considered as the 
powers of some one number, arbitrarily chosen, which is called 
the base of the system, and the exponent of that power of the 
base which is equal to any given number, is called the loga- 
rithm of that number. 


Thus, if a be the base of a system of logarithms, N any 
number, and x such that 
N ='a* 


then z is called the logarithm of N in the system whose base 
is a. 

The base of the common system of logarithms, (called from 
their inventor “ Briggs’s Logarithms”), is the number 10. 
Hence since 


(loy°= 1 


, 0 is the logarithm of 1 __ in this system. 
(0)'= 10 ,1 ee cae a ee 
{RO} = vib Onis! 2 Wie elt ce Cee 
(10)s= 1000 , 3 3008: 6 
(10)*= 10000 , 4 10000 ——-———— 
&e.. =). &e: Sicviny.o; eaeee: with 


From this it appears, that in the common system the loga- 
rithms of every number between I and 10,is some number be- 


ANALYTICAL PLANE TRIGONOMETRY. 79 


tween 0 and 1,7. e. is 1 plus a fraction. The logarithm of every 
number between 10 and 1V0, is some number between 1 and 
2,2. e. is 1 plus a fraction. The logarithm of every number 
between 100 and 1000, is some number between 2 and 3, 7. e. 
is 2 plus a fraction, and so on. 

In the common tables the fractional part alone of the loga- 
rithm is registered, and from what has been said above, the 
rule usually given for finding the characteristic, or, indea, i. e. 
the integral part of the logarithm will be readily understood, 
viz. The index of the logarithm of any number greater than 
unity is equal to one less than the number of integral figures in 
the given number. Thus, in searching for the logarithm of 
such a number as 2970, we find in the tables opposite to 2970 
the number 4727564; but since 2970 is a number between 
1000 and 10000, its logarithm must be some number between 
3 and 4, 2. e. must be 3 plus a fraction; the fractional part is 
the number 4727564, which we have found in the tables, affix- 
ing to this the index 3, and interposing a decimal point, we 
have 3,4727564, the logarithm of 2970. 

We must not, however, suppose that the number 3.4727564 
is the exact logarithm of 2970, or that 


9970 = (1 0) 3.4727564 


accurately. The above is only an approximate value of the 
logarithm of 2970; we can obtain the exact logarithm of very 
few numbers, but taking a sufficient number of decimals, we 
can approach as nearly as we please to the true logarithm, as 
will be seen when we come to treat of the construction of 
tables. 

It has been shown that in Briggs’ system the logarithm of 
1 is 0, consequently, if we wish to extend the application of 
logarithms of fractions, we must establish a convention by 
which the logarithms of numbers less than 1 may be repre- 
sented by numbers less than zero, i.e. by negative numbers. 


Extending, therefore, the above principles to negative ex- 
ponents, since 


= or (10)’"=0.1, —1 isthe logarithm of.1 in this system 
! 2 
Foo °F (10)"=0.01, —-2 —————_ 01 
1 
sel 5 ——(. eit —————— 001 
5900 OF (10)-*=0.001.—3 
1 
——— 10)—"=0.0001,--4 ——-—————__ .0001. 
700007 « )-‘=0.00 4 


&c. &c. 


80 ANALYTICAL PLANE TRIGONOMETRY. 


It appears, then, from the convention, that the logarithm of 
every number between 1 and .1l, is some number between 
0 and —1; the logarithm of every number between .1 and 
.01, is some number between — 1 and— 2; the logarithm of 
every number between .01 and .001, is some number between 
— 2 and —3; and so on. 

From this will be understood the rule given in books, of ta- 
bles, for finding the characteristic or index of the logarithm of 
a decimal fraction, viz. The index of any decimal fraction is a 
negative number, equal to unity, added to the number of zeros 
immediately following the decimal point. 'Thus, in searching 
for a logarithm of the number such as .00462, we find in the 
tables opposite to 462 the number 6646420 ; but since .00462 
is a number between .001 and .0001. its logarithm must be 
some number between — 3 and —4, 2. e. must be —3 plus a 
fraction, the fractional part is the number 6646420, which we 
have found in the tables, affixing to this the index — 8, and in- 
terposing a decimal point, we have — 3.6646420, the loga- 
rithm of .00462. 


General Properties of Logarithms. 


Let N and N’ be any two numbers, z and z’ their respec- 
tive logarithms, a the base of the system. Then, by def. (2), 


N=a¥- - - - - - (1) 

Nisa vie) 2) <p) = 1(2) 

I. Multiply equations (1) and (2) together, 
N NS=a* a 
— gst 


. by def. 2, 2+2’ is the logarithm of N N’, that is to say, 


The logarithm of the product of the two or more factors is 
equal to the sum of the logarithms of these factors. 


II. Divide equation (1) by (2), 
RT 
N’ ak 

—q*-* 


def. (2), x--z' is the logarithm of ae that is to say, 


ANALYTICAL PLANE TRIGONOMETRY. 81 


The logarithm of a fraction, or of the quotient of two num- 
bers, is equal to the logarithm of the numerator minus the loga- 
rithm of the denominator. 


III. Raise both members of equation (1) to the power of n. 
N2 =q@"* 


by def. (2), x is the logarithm of N *, that is to say, 


The logarithm of any power of a given number is equal to 
the logarithm of the number multiplied by the exponent of the 
power. 


IV. Extract the n” root of both members of equation (1). 
ntaa 
1 
by def. (2), is the logarithm of No that is to say, 
The logarithm of any root of a given number is equal to the 
logarithm of the number divided by the index of the root. 


Combining the two last cases, we shall find, 
m mx 
Na = Oy 


whence, sl is the logarithm of N». 
It is of the highest importance to the student to make him- 
self familiar with the application of the above principles to al- 


gebraic calculations. The following examples will afford a 
useful exercise : 


Ex. 1. log. (a. b.c. d....)=log. at+log. b+log. c+tlog.d+... 


b 
Ex. 2. log. —-) =log. atlog. b+log. c—log. d—log e. 
Ex. 3. log. (a™b" cP. ...)=m log. a+n log. b+p log.c... 

m }n 
Ex. 4. log. (- % ) =m log. a+n log. b—p log.c... 


Ex. 5. log.(a’—2z*)=log.(a+2z).(a—z)=log.(a+x) +log.(a—z) 


a2 1 1 
Ex. 6. log. Vq?—z’ ages log. (a+z) > (a—x) 


@ 


82 ANALYTICAL PLANE TRIGONOMETRY. 


Ex. 7. log. (a*4/a°)=log. a’+ Flog. a’=8 log. a+ log. a 


15 
= _ log. a 


Ex, 8. log. V(@—2)"= = log. (a—2)+— log. (@’-+ar-+2') 


= = slog. (a—zx) +log. (a+x2+z)+log. (a+r—z)} 
| where z?=az 
Ex. 9. log. /q?ta? = = flog (a+xz+z)+log. (a+rz—z)}, 
where z?=2axr 
V HZ, 4 
Ex. 10. log. facie ieee, (a—zx)—3 log. (a+z)} 


Let us resume the equation, 
NN == a 

1°. If a>1, making z=0, we have N=1; the hypothesis 
x=1 gives N=a. As z passes from 0 up to 1, and from 1 up 
infinity, N will increase from 1 up to a, and from a up to in- 
finity ; so that x being supposed to pass through all interme- 
diate values, according to the law of continuity, N increases 
also, but with much greater rapidity. If we attribute negative 


iB 1 4 
values toz. wehave N=a *, or N= —— Here, as x increas- 
= 


es, N diminishes, so that z being supposed to increase nega- 
tively, N will decrease from 1 towards 0, the hypothesis 
z=o gives N=0. - 


1 
2°. If a<, put a= he where b>1, and we shall then have 


1 clade : \. 
N= jx GS N=0* , according as we attribute positive or nega- 


tive values to z. We here arrive at the same conclusion as 
in the former case, with this difference, that when z is po- 
sitive N<1, and when z is negative N>1. 


3°. If a=1, then N=1. whatever may be the value z. 


From this it appears, that, 


1. In every system of logarithms the logarithm of 1 1s 0, and 
the logarithm of the base is 1. 


ANALYTICAL PLANE TRIGONOMETRY. 83 


II. If the base be >1, the logarithms of numbers >1 are po- 
sitive, and the logarithms of numbers<1 are negative. The 
contrary takes place if the base be<l. 


III. The base being fixed, any number has only one real log- 
arithm ; but the same number has manifestly a different loga- — 
rithm for each value of the base, so that every number has an infi- 
nite number of real logarithms. Thus, since 9°=81, and 3*=81, 
2 and 4 are the logarithms of the same number 81, according 
as the base is 9 or 3. 


IV. Negative numbers have no real logarithms, for attribut- 
ing to x all values from —a@ up to+ @, we find that the corres- 
ponding values of N are positive numbers only, from 0 up 
to + @. 

The formation of a table of logarithms consists in deter- 


mining and registering the values of z which correspond to 
=, 2,)3,. 7. 5. Ingthe.equation, 


N=a* 
If we suppose m=a*, making 
L=0, a, 2a, Ba, 4a, 5a, - - - - - logarithms. 
y=1, m, m’*, m*, m‘, mm’, - - - - = numbers. 


the logarithms increase in arithmetical progression, while the 
numbers increase in geometrical progression ; 0 and 1 being 
the first terms of the corresponding series, and the arbitrary 
numbers « and m the common difference and the common ratio. 

We may, therefore, consider fhe systems of values of x and 
y, which satisfy the equation N=a*, as ranged in these two 
progressions. 

In order to solve the equation 

c= 

where c and a are given, and where z is unknown, we equate 
the logarithms of the two members, which gives us 


log. c=z log. a 
Whence, 
gules: c 
log. a 


To determine the value of x in the equation 
Aa*® + Bat—? + Cat—* + ve iP 


ByosG 
a (A+ to ee aaa, ee 


we have 


7 
sat 


84 ANALYTICAL PLANE TRIGONOMETRY. 
_» = Or, + : r 
° * | Qa* =P 
log. P—log. Q 
2 
log. a 


_If we have an equation a? = b, where z depends upon an 
- unknown quantity z, and we have 


z=Aa® + Bo®—-14.... 


% 


log. b 


“Since z=7 ~ = some known number, the problem depends 


upon the solution of the equation of the n™ degree. 
K= Aa +-Batc hp 29. 
For example, let 
2, v#—5r+4 
4(3) =9 
Hence, 
2 9 
(’— 52+-4) log. (=) =log. 7 


v’— 5e+4 = — 2 
an equation of the second degree, from which we find « = 2, 
z= 3. 
To find the value of x from the equation 


bx 2 cm TRA 
Taking the logarithms of each member, 
(n— <) log. b=mx log. c+ («—p) log. f 
Or, 
(m log. c+log. f) 7° — (n log. b-+p log. f) «+a log. b=0 


a quadratic equation, from which the value of « may be de- 
termined. 


In like manner, from the equation 
cm™x — ghrx—1 
we find 
_ log. a— log. b 
*~"m log. c—n log. b 
Equations of this nature are called Exponential Equations. 


Let N and N + 1 be two consecutive numbers, the differ- 
ence of their logarithms, taken in any system, will be 


N+1 1 
log. (N+1) —log. N=log. (“) =log. (1+ ne 


a quantity which approaches to the logarithm of 1, or zero, in 


* 


- 


ANALYTICAL PLANE TRIGONOMETRY. "85 


proportion as - decreases, that is, as N increases. Hence it _ 


appears, that 


The difference of the logarithms of two consecutive numbers 
zs less in proportion as the numbers themselves are greater, 

When we have calculated a table of logarithms for 
any base a, we can easily change the system, and calculate 
another table for a new base b. 


Let c=0*, z is the log of c in the system whose base is b; 


_ Taking the logs. in the known system, whose base is a, we have 


» lox®e 1 
ce 7 j =log.c (saa) ered (A) hence 


The log. of cin the system whose base is b, is the quotient 


arising from dividing the log. of c by the log. of the new base b, 


“both these last logs. being taken in the system whose base is a. 
In order .*. to have x the log. of c in the new system, we 


1 ] 
must multiply log. ¢ by are, this last factor loa Is cOn+ | 


stant for all numbers, and is ae the Modulus ; that is to say, 
if we divide the logs. of the same number c taken in two sys- 
tems, the quotient will be invariable for these systems, what- 
ever may be the value of c, and will be the modulus, the con- 
stant multiplier which reduces the first system of logs. to the 
second. 

If we find it inconvenient to make use of a log. calculated to 
the base 10, we can in this manner, by aid of a set of tables 
calculated to the base 10, discover the logarithm of the given 
number in any required system. 


For example, let it be required, by aid ef Briggs’ tables, to 
i) 
find the log. of < in a system whose base is 7 
Let x be the log. sought, then by (A) 
2 


log. 3 

x= 7 

log. 7 
_log. 2—log. 3 
~ log. 5 — log. 7 


g* 


86 ANALYTICAL PLANE TRIGONOMETRY. 


Taking these logs. in Briggs’ system, and reducing, we find. 
— 0.17609125 


— —0.14612804 
2 5 
== 1.2050476 = log. 3 t° base 7° 


3 
Similarly, the log. of <, in the system whose base is =, is. 


2 
__ log. 2—log. 3 
“Tog. 3— log. 2 
=—] 
which is manifestly the true result ; for in this case the gene- 
' 2 Beek Pew al os 
ral equation N=a* becomes = =(2)—(2) , and & is evi- 
dently = —1. 
In a system whose base is a, we have 
wi is 1 °8: n 


for, by the definition of a log. in the equation n=a*, z is the 
log. n. 
In like manner, 
: h 
at Se doe (hye at log. n. 


EXAMpLes FoR EXERCISE. 


Ex. 1. Given 2* +2* =12 to find the value of z. 
Ans. z=1.584962, or z=log. (—4)+log. 2. 
Ex. 2. Given z+y=a, and m@~¥)=n to find z and y. 
Ans. z=}{a+log. n+log.m} and y=}}a—log. n+log. m}. 
Ex. 3. Given m* n* =a, and hr=ky to find z and y. 
‘ z=log. a (log. m+log. n) 
ns. 


and y=—_log. a~- (log. m+-log. n). 


To find the logarithm of any given number. 
Let N be any given number whose logarithm is z, in a sys- 
tem whose base is a; then 
aX =N and a@=N?; 


hence, by the exponential theorem, we have from the last 
equation 


1+Aaz+A*~-+.., =1+A,2+A 22 +... 
1.2 1.2 


“ 


ANALYTICAL PLANE TRIGONOMETRY. 87 


and equating the coefficients of z, we get Ar=A,; hence 
A, _ (N—1)—H(N—1)'+(N—1)—... 

A (a—1)—4(a—1)*+3(a@—1)—. . 
because A=(a—1)—}(a—1)?+1(a—l)’"— . . . . in the 
expansion of a*4. 
and A, =(N—1)—}(N—1)? +34(N—)"— .. . . in the 
expansion of N?. 


& 


To find the logarithm of a number in a converging series. 


We have seen that if aX =N?, then 
pI) —H(N 1) 4 (N1 1) (Nt 
(a—1)—}(a— 1 +i(a— 1)—3(a— y+ 
Now the reciprocal of the denominator is the modulus of 
the system ; and, representing the modulus by M, we have 
z=log. N'=M$(N'—1)—3(N'’—1)?+3(N*—1)°—1 (N'—1)‘+ 
Put N'=1+7; then N'—l=n2, and we have 
log. (1-+-n) =M(+n—in’+1in’—in‘+in'—...3 
Similarly log. (l—n)=M(—n—3n*—1n’—1n'‘— 1n'—. . .) 
*. log. (1+7)—log. (l—2)=2M(n-+-in'+in'+in'+...) 


1 
or Jog. OM (ntin'+ n't int. 9. 


1 
Pot 2=. 2; then 14-n=2 = 1— 
BP aPLL 


2P nd it” ‘att ot Oe 


Sra | 


Sep OnE | os 
~ Q2P+1’ I—xn P 


consequently 
1 1 1 1 
log.(P+1)—log.P=2M SP41* 3@PH1) BPH) 


‘ 1 1 ee 

sfaamAR tH aalop Eat ai | aP41+ 3@P41)* 5RP+I) 

Hence, if log. P be known, the log of the next greater num- 
ber can be found by this rapidly converging series. 


To find the Napierian logarithms of numbers. 


In the preceding series, which we have deduced for log. 
(P+1), we find a number M, called the modulus of the sys- 
tem; and we must assign some value to this number before 
we can compute the value of the series. Now, as the value 
of M is arbitrary, we may follow the steps of the celebrated 
Lord Napier, the inventor of logarithms, and assign to M the 
simplest possible value. This value will therefore be unity ; 


and we have 


88 ANALYTICAL PLANE TRIGONOMETRY. 


; 1 1 1 
log. (P+1)=log. P+2 sP+i? 3@P41)t 5@PHIy 


Expounding P successively by 1,2,3,4, &c., we find 


ip cae aga 
_—_— _— eooe Po of 1 
Ppeathatet )= .6931472 


2=log. 1+2( vtee 


1 1 1 1 
log. 3=log. 242(stesteptagt +) = 10086128 
log. 4=2log.2 - - - °- = =1,8862944 
1 1 1 ] 
log. 5=log. 442(S tan tag tag t) =1.6004379 
log. 6=log. 2+log.-8 - = - = + =1.7917595 
1 1 1 
los. Y=log. 6+2 (stage teagt~)=1 0450101 
log. 8=log. 2+log. 4, or 3 log.2. - - =2.0794415 
log... 9=2 log. 3 - - - - - =2.1972246 
log. 10=log. 2+log.5 —- - - - =2.3025851 


In this manner the Napierian logarithms of all numbers may 
be computed. 


To find the common logarithms of numbers. 


Let a} = N and.bY= N; then we have 
« = log. N to the base a, or z=log. ,N 
y = log. N to the based, or y=log. ,N 
hence, log. ,N=log. aby =log. ab (Gen. properties logarithms.) 
é& 2=y log. ,b 


and y= loeb 


and by means of this equation we can pass from one system of 
logs. to another, by multiplying 2, the log. of any number in 
the system whose base is a, by the reciprocal of log. bin the 
same system ; and thus we shall obtain the log. of the same 
number in the system whose base is 0. 


Let the two systems be the Napierian and the common, in 
which the base of the former is €=2.718281828... and the 
base of the latter is b5=10, the base of the common system of 
arithmetic ; then we have 6=10, and a=e«=2.718281828... 


ANALYTICAL PLANE TRIGONOMETRY. 89 


and consequently if N denote any number, we shall have 
log. wN=55 7108 eN ; thatis, 

nap. log.N nap. log.N 

nap.log.10 .2.3025851 

and the modulus of the common system is, therefore, 

1 

M=9 "3025851 

Hence, to construct a table of common logarithms, we have 

1 1 
3@P +1) *5@P+1)" 
Expounding P successively by 1,2,3, &c., we get 


com. log N= = 343429448 Xx nap. log.N; 
=.43429448 .. 2 M=.86858896 


1 
log(P+1)=logP +.86858896 spit 


log. 2=.86858896( 5+ a+ ; pas) 


5.3° 
=.86858896 X .6931472 - ‘ - =,.38010300 
lie She ak 
log. 8—log.2-+.86858890( = + 5554-5,+. A )=47213 
log. 4=2log:2 - - sd ererg™ - = =.,.6020600 
log. 5=log. 42=log. 10—log 2=1—log.2 - =.6989700 
log. 6=log. 2+log.3 = 9 - = =e) -=.7781513 
eee | 1 | 
log. 7=log.6-+ 968588963 +59 + sagt ~ =,8450980 
log. 8=log.2%=8log.2  - - -  - =.9030900 
log. 9=log.3* =2 log.3—- - . - =.9542426 
Lg. 10 cry Sh mig oe Be) oct nd =a =1,0000000 
&ce. &c. 


, 1 
Since log. "= 2M (nin? + In? +1n7 +...) 


1+n te _P—l 
let ——=P ; then 1+n=P (1—n) or 2=Pq 


+. log. P=2M } eaiteg (aay ) +5(pq) +. ‘3 


and thus we have a series for computing the logs. of all num- 
bers, without knowing the log. of the previous number. 


90 ANALYTICAL PLANE TRIGONOMETRY. 


EXxampues 1N LoGARITHMS. 


(1.) Given the log. of 2=0.3010300, to find the logs. of 25 
and .0125. 


100 10? 
Here 20 OE 
therefore log. 25=2 log. 10—2 log. 2=1.3979406 
; 125 1 ] 
Again .0125 =T00007 80 10x22 


.“. log. .0125=log. 1—log. 10—3 log. 2=—1—3 log. 2 
=2.0969100 
(2.) Calculate the common logarithm of 17. 

Ans. 1.2304489. 

(3.) Given the logs. of 2 and 3 to find the logarithm of 12.5. 
Ans. 1+2 log. 3—2 log. 2. 

(4.) Having given the logs. of 3 and .21, to find the loga- 

rithm of 83349. 

Ans. 6+2 log. 34-3 log. .21. 


On ExponentTIAL EQuatTIons. 


An exponential equation is an equation in which the un- 
known quantity appears in the form of an exponent or index ; 
thus, the following are exponential equations : 


x x 
a Deak Ge eP ian, Cua =: ay OCG: 


When the equation is of the form a* = b, or a? =c, the 
value of x is readily obtained by logarithms, as we have al- 
ready seen above. But if the equation be of the form 
x* = a, the value of xz may be obtained by arule of approz- 
imation, as in the following example : 


Ex. Given «* = 100, to find an approximate value of z. 
The value of «is evidently between 3 and 4, since 33 = 27 


and 4* = 256; hence, taking the logs. of both sides of the 
equation, we have 


ANALYTICAL PLANE TRIGONOMETRY. 91 
«log. c=log. 100=2 


First, let 2; = 3.5; then 
3.5 log. 3.5= 1.9042380 
true no. = 2.0000000 


Second, let x2 = 3.6; then 
3.6 log. 3.6= 2.0026890 
true no. = 2.0000000 


error =—.0957620 | error = +.0026890 


Then, as the difference of the results is to the difference of 
the assumed numbers, so is the least error to a correction of 
the assumed number corresponding to the least error ; that is, 

098451. ;.1,: : .002689 :. .00273; 
hence t=3.6—.00273=3.59727, nearly 


Again, by forming the value of «* for z=3°5972, we find the 
error to be—.0000841, and for x=3.5973, the error is + 
0000149 ; 


hence, as .000099 : .0001 : : .0001 : .0000151 ; 
therefore <=3.5973—.0000151=3.5972849, the value nearly. 


EXxaAmMPLes For PRACTICE. 


(1.) Find x from the equation 2* = 5 Ans. 2.129372. 
(2.) Solve the equation z* = 123456789 Ans. 8.6400268, 
(3°) Find z from the equation 2* = 2000. Ans. 4.8278226. 


Since the properties of logarithms afford great facilities in 
performing complicated arithmetical operations upon large 
numbers, it becomes desirable to have the logarithms of sines, 
cosines, tangents, &c. computed and arranged in tables; but 
most of these numbers being less than unity, their logarithms 
would, of course, be negative. To avoid this inconvenience, 
all the trigonometrical functions calculated in the manner 
explained in Chap. 1V, are multiplied by a large number, and, the 
operation being performed upon all, their relative value is not 
altered. This number may, of course, be any whatever, pro- 
vided it be so large, that, when the numerical values of trigo- 
nometrical quantities are multiplied by it, their logarithms may 
be positive numbers. 

The number employed for this purpose in the common tables 
is 10000000000 or 10”, which is usually represented by the 
symbol R. 

The sine of 1’, as computed above, is 

Sin. 1/=.0002908882 .... 


92 ANALYTICAL PLANE TRIGONOMETRY. 


a number much smaller than unity, and whose logarithm 
would consequently be negative. 

When multiplied by 10° it becomes 

== “2908882 22° 
a number whose logarithm is 6.4637261, and consequently we 
find in our tables, log. sin. 1'=6.4637261. 

A table constituted upon this principle is called a Table of 
Logarithmic Sines, Cosines, Tangents, &c. and by this nearly 
all the practical operations of trigonometry are usually per- 
formed. 

It is manifest, from these remarks, that before we can apply 
formule deduced in the preceding chapters to practical pur- 
poses, we must transform them in such a manner as to render 
the several trigonometrical quantities identical with those re- 
gistered in our tables. The sines, cosines, &c. we have hith- 
erto employed, are called Trigonometrical qnantities calculated 
to a radius unity ; those registered in the tables, Trigonome- 
trical quantities calculated to radius R. 

The problem to be solved therefore is 

To transform an expression calculated to a radius unity, to 
another calculated to a radius R 

Let us represent sin. 4 to radius unity by m. 


me 2 HF by n. 
Then the relation between them is 
~ on =\“Rien 
rege 7 
id Reilly 


and so for all the other trigonometrical quantities. 

Hence, in order to transform an expression calculated to ra- 
dius unity, to another calculated to radius R, we must divide 
each of the trigonometrical quantities by R. 

If any of the trigonometrical quantities enter in the square, 
cube, &c. these must of course be divided by R’? R’, &e. ...... 

As observed above, R may be any given number whatever, 
the number usually employed in the ordinary tables being 10”, 
and therefore 

log. R = 10 
Take as an example such an expression as 

asin. 6 = 6 tan.’ o 
in order to reduce this to an expression which we can com- 
pute by our tables we must, according to the above rule, di- 
vide each of the trigonometrical quantities by the proper 
power of R: the expression then becomes 

sin. 4 tan.” 


—— ee) 


eet eats 45 ic 


ANALYTICAL PLANE TRIGONOMETRY. 93 


Or, clearing of fractions, 
aR sin. é = b tan. o 
Or, 
log. a + log. R + log. sin. 6 = log. b + 2 log. tan. o » 
an expression which may be calculated by the tables. 
If the expression calculated to radius unity be of the form 
sin. 6 


sin. o 
it requires no modification, for if we divide both terms of the 
sin. 


6 
fraction by R, we shall not alter its value. 


sin. @ 

We need not prosecute this subject farther, as numerous ex- 

amples of these transformations will occur at every step in the 
succeeding chapters. 


CHAPTER VI. 


ON THE SOLUTION OF RIGHT-ANGLED TRIANGLES. 


Every plane triangle being considered to consist of 6 parts, 
the three sides, and the three angles, if any of these three parts 
be given, we can, in general, determine the remaining parts 
by trigonometry. 

In right-angled triangles, the right angle is always known, 
and therefore any two other parts being given, we can, in ge- 
neral, determine the rest. We shall thus have five different 
cases. 

1. When one of the acute angles and the hypothenuse is 
given. 

2. When one of the acute angles and a side is given. 

3. When the hypothenuse and one side is given, 

4, When the two sides are given, 

5. When the two acute angles are given. 

Let ABC be a right-angled triangle, C the right ngle, 


Let the sides opposite to the angles A and B 
be denoted respectively by a, 6, and let the hy- 
pothenuse be called c. 
A c 
Case 1, Given A, c, required B, a. b. 
Since C is a right angle 
A+B=90° 
es B=90°—A whence B is known - - -(3) 
By Chap. III. prop. 1, | 
G90 Sits A 
9 


? ' 


94 ANALYTICAL PLANE TRIGONOMETRY. 


Adapting this expression to computation by the tables. 
) log. a=c sin. A 
R 


ae log. a=log. c+log. sin. A—log.R, whence a is known - (2) 
In like manner, 
bac CO A 


“. log. b=log. c+log. cos. A—log. R, whence b is known (8) 

If B, c are given, and A, a, b required, we shall have pre- 
cisely in the same manner, 

A=90°— Beis wee nw, ive = (4) 

log. a=log. c+log. cos. B—log. R- - - - - - (5) 

log. b=log. c+log. sin. B— log. R- - - - - - (6) 


Case 2. Given A, a, required B, 8, c. 
B=90°—A, whence Bisknown - - - - - (7) 
b=a7e0tenX 

Adapting the expression to computation by the tables. 


R 
log. b=log. a+log. cot. A—log. R, whence b is known (8) 
Again, 
a=c sin, A 
a 
sin. A 
- Adapting it to computation, 
c=kR . —% 
sin. A 


. log. c=log. R+log. a—log. sin. A, whence c is known (9 
If A, b, be given, B, a, c, we shall have in like manner, ) 


Pi 


00) on ae en tere Peete Loe wa ee (10) 

log. a=log. b+log. tan. A—log.R - - . - - (11) 

log. e=-loge Kh +-log. b-— log. cos: Acts ti (12) 
If B, b be given, and A, a, c required, 

A=90° —B FR a ee il ee a Hae (13) 

log. a=log. b+log. cot. B—log. R - - - . . (14) 

log. c=log. R+log.b—log. sin. Bo . - - = - (15) 
If B, a be given, and A, 8, c, required. 

Fe On TS mien om scons) ot Vo eran Ser. Te (16) 

log. b=log. at+log. tan. B—log. R - - - . | (17) 

log. e=log. R+log.a—log.cos.B - - - = . (18) 


ANALYTICAL PLANE TRIGONOMETRY. 95 


Case 3. Let a, c be given, required b, A, B. 


b’=c? —a’ 
=(c+a) (c—a) 
“. 2 log. b=log. (ec+a)+log. (c—a) whence d is known - - (19) 
sin. A=— 


Adapting the expression to computation. 
‘sin. A 


R —= 
.. log. sin. A=log. R+log. a—log. c, whence A is known, (20) 
So also, 
cos. Ba 


——_— 


a 
Cc 


c 
.“. log. cos. B=log. R+log. a— log. c, whence B is known, (21) 
Ifb, c be given, and a, A, B required, we shall have 


2 log. a=log. (c+b)+log. (c—b) - - - - - (22) 
log. cos. A=log. R+log b—log.c + - - - - - (28) 
log. sin. B=log. R+log. b—log.c - - - + - - (24) 


Case 4. Let a, 6 be given, required A, B, c. 
tan. A= 
b 


Adapting the expression to computation. 
ii fatame Ad -@& 


“piece ane : 
.“. log.tan. A=log. R+log. a— log. 6, whence A is known, (25) 
So also, 


tan. Bd 
wR a 
“. log. tan. B=log. R+log. b — log. a, whence B is known, (26) 
c=Vq'+0*, whence cis known, - - - - (27) 


Case 5. Given A, B, required a, }, c. 


It is manifest that this case does not admit of a solution, for 
any number of unequal similar triangles may be constructed, 
having their angles equal to the angles A, B, C. 


We shall conclude this chapter by giving some numerical 
examples. 


Example 1. Given A=26° 41’ 6”, c=6539.76 yards, re- 
quired a. ; 


96 ANALYTICAL PLANE TRIGONOMETRY. 


Then by (2). 
log. a=log. c+log. sin. A— log. R. 
By the tables log.c= 3.8155618 
log. sin. A= = 9.6523286 
13.4678904 
log. R= 10. 


log.a= = 3.4678904 
The number in the tables corresponding to the logarithm 
3. 4678904 is found to be 2936.91. 
a=2936.91 yards. 
In like manner, the side b may be determined, if required. 


Example 2. Given c=6539.76 yards, a=2936.91 yards, re- 
quired b, A, B. 
By (19). 
™ c+a=9476.67 
2 log. b=log. (c+a)+log. (ec —a) ¢ a= 3602.85 


By the tables, 
wi bes log. (cta)= 3.9766557 
log. (c—a)= 3.5566462 
2 log. b= '7.5333019 
log. b= 3.7666509 
The number in the tables corresponding to the logarithm 
3. LE is 5843.2 . 
b=5843.2 yards. 


To aeterntine A we have (20). 
log. sin. A=log. R+log. a— log. ¢ 


By the tables, log. "a= 3.4678904 7 
log. (R=>,10. 
13.4678904 


log. c=  3.8155618 


log. sin. A= 9.6523286 
On referring to our tables, we shall find that the angles 
whose logarithmic sine is 9.6523286 is 26° 41’ 6”, which is 
consequently the value of A. 
A being known, B is determined at once by subtracting the 
value of A from 90°, or B may be determined independently 
of A by (21). 


a 


log. cos. B=log. R+log. a — log. c. 


— ile é bl 


ANALYTICAL PLANE TRIGONOMETRY. 97 


Example 3. Let a, 6,in the last number be given, and c 
required. 

Then by (27). . 

c= V+) or C= +0? 

The calculation in this case is not so simple, for the quan- 
tity under the radical cannot be easily adapted to logarithmic 
calculation. 

We have, log. a?=6.9357808 .°. a’= 8625400 

log. b°=7.5333019 .. b’= 34143000 


c’= 42768400 
log. c?=7.6311230 
log. c =3.8155615 
c =6539.76 


Example 4. Given c=6512.4 yards, b=6510.6, to find A. 
By (28). 
log. cos. A=log. R+log. b— log. c 
Now, loo’ R= 10. 
log. b = 3.8136210 


13.8136210 
log. c= 3.8187411 


log. cos.. A= -9:9998799 
A= 1° 20’ 50” 


Upon inspecting the tables that are calculated to seven 
places of decimals only, it will be seen that, when the angles 
becomes very small, the cosines differ very little from each 
other. The same remark applies, of course, to the sines of an- 
gles nearly 90°. In cases, therefore, where great accuracy 
is required, we may commit an important error by calculating 
a small angle from its cosine, or a large one from its sine. We 
must consequently endeavor to avoid this, by transforming 
our expression by help of the relations established in chapter _ 
first and second. 

In the example before us, A is a small angle which has been 
calculated from its cosine ; we must therefore, if possible, cal- 
culate this angle by means of its sine, or some other trigono- 
metrical function. 


Now, by formula (7 ), chap. II. we have generally 


ames. + 
sin. = 4/1 cos. A 


2 


Q* 


. al 
“ , 


93 ANALYTICAL PLANE TRIGONOMETRY. 


a Koc: me 
In the present case, cos. a substituting this in the 


above equation, 
ig gibdg fe? 
sin. ews Segond 
log. sin. =} log. (ce —a) — log. 2 a+log. R. 
From which we find, 


A 
. 9 40 24" 
And $ Aa OU aa 


Instead of 1° 20’ 50”, as obtained by the former process. 


No angle which is nearly 90° ought to be calculated from 
its tangent, for the tangents of all large angles increase with so 
much rapidity, that the results derived from the column of 
proportional parts found in the tables cannot be depended on 
as accurate. 


CHAPTER. VIL 


ON THE SOLUTION OF OBLIQUE ANGLED TRIANGLES. 


Six different cases present themselves. 


1. When two angles and the side between them are given. 

2. When two angles and the side opposite to one of them are 
given. 

. When two sides and the included angle are given. 

. When two sides and the angle opposite to one of them are 
given. 

. When the three sides are given. 

. When the three angles are given. 


we C2 


S> Or 


Let A, B, C be a plane triangle. 

Let the angles be denoted by the large letters 
A, B,C, and the sides opposite to these angles by 
the corresponding small letters a, 8, c. 


Case 1. Given A, B, ¢, required C, a, b. 
Since A+B+C=180° 

C=180°—(A+B,) whence C is known. 

© being thus determined, we have, by chap. III. prop. 2, 


A B 


ANALYTICAL PLANE TRIGONOMETRY, 99 


a sinA 
ec sin.C 
sin. A ¥ 


fy" sin. G 
An expression which is in a form adapted to computation by 
the tables. 
“. log. a=log. c+log. sin. A—log. sin. C, whence a is known. 
Again, 


Og sin. B 

cc. sin,C 
sin. B 

Lyset: sin. C 


*, log. b=log. c+log. sin. B—log. sin. C, whence 3 is known. 


If any other two angles and the side between them be given, 
we may determine the remaining angle and sides in a manner 
precisely similar. 


Case 2. Given A, B, a, required C, 3, ec. 


Since eas Lae 
C=180°—(A+B,) whence C is known. 


: b sin. B 
Again, asin. A 
Abe ciisisle. 
arcane PE 


“. log. b=log. a+log. sin. B—log. sin. A, whence 0 is known. 
Also, C being known, 


c ‘sin. C 

a sin, A 
mo sins 

oe mn. 1A 


. log. c=log. a+log. sin. C—log. sin. A, whence c is known. 

Tf any two other angles and the side opposite to one of them 
are given, the remaining angle and sides may be determined 
in a manner precisely similar. 

Case 3. Given a, b, C, required A, B, c. 

By prop. 3, chap. III. 
A+B - 

2 a+b 
Die GLABINgL Ip Ue FAC hatiamorye ed AD 
tan. 3 


tan. 


¥, ; ¥ 
100 ANALYTICAL PLANE TRIGONOMETRY. 
Now, A+B+C=180° 


A 
Substituting this value of tan. 


cot. ay a+b 

VURSA= By aep 

tan. 5) | 
A—Be.a=p 


. ——_ =— cot. — 
we 2 PTT oy 


log. tan. 5 =log. (a—b) +log. cot. —log. (a+d) 
And we can thus calculate the value of the angle " from 
our tables; let the angle thus found be called 9 ... A-~B=2 9. 
Now A+B=180°—C 
And . A—B=2 9 
.. adding and subtracting A=90°+9—~ 


C 
B=90°—(o +5) 


The angles A and B will thus become known, and, these 
being determined, we can find the side c from the relation, 
ec sin. C 
a sin. A 
sin. C 
sin. A 
log. c=log. at+log. sin. C—log. sin. A 


c=a 


If a, c, B, or b, c, A be given, the remaining angles and side 
may be determined in a similar manner by aid of the formula 
(j) in chap. III. 


¥. piled ia 


ANALYTICAL PLANE TRIGONOMETRY. 101 


Case 4. Given a, 6, A to determine B, C, ec. 
sin. Bb 
sin. A a 


sin. B—=sin. A. 2 
a 


‘. log. sin. B= log. sin. A+log. ’— log. a, whence c is known. 
B being known, C = 180° —(A+B,) whence C is known. 


Cc sin. C 
C being known, —=—= — 
8 7a sin. A 


: sin. C 
Caton 
“. log. a= log. sin. C — log. sin. A, whence 0 is known. 


If any two other sides and the angle opposite to one of them 
be given, the remaining angles and side may be determined 
in a manner precisely similar. 

It must be remarked. that, in the above case, we determine 
the angle B from the logarithm of its sine; but since the sine 
of any angle, and the sine of its supplement are equal to one 
another, and since it is not always possible for us to ascertain 
a priort whether the angle B is acute or obtuse, the solution 
will be sometimes ambiguous. 

In fact, two different and unequal trianvles in! 
may be constructed, having two sides and 
the angle opposite to one of those sides in p 
one triangle, equal to the corresponding sides 
and angle of the other; one of these trian- 
gles will be obtuse-angled, and the other 
acute-angled, and the angles opposite the re- » 9 
maining given sides in each will be supple- 
mental. 

Thus let ABC, be a plane triangle. 

With centre C and radius equal to CB describe a circle cut- 
ting AB in D. 

Join CD. 

Then it is manifest that the two unequal triangles CBA, 
CDA, have the two sides CB, CA of the one, equal to the two 
sides CD, CA of the other, and the angle A, opposite the equal 
sides CB, CD, in each, common. 

It is manifest from this, that it is impossible to determine ge- 
nerally, from the data of this case, which of the two triangles 
is the solution of the problem. There are certain considera- 
ear ees by which the ambiguity may sometimes be re- 
moved. 


102 ANALYTICAL PLANE TRIGONOMETRY. 


1. If the angle be obtuse, then both of the remaining angles 
must be acute, and the species of B will be determined. 

2. If the given angle be acute, but the side opposite the 
given angle greater than the given side opposite the required 
angle, then the required angle is acute. For since in every 
triangle the greater side has the greater angle opposite to it, 
and since the side opposite to the given angle, which is acute, 
is greater than the side opposite to the required angle, it fol- 
lows, a fortiori, that the required angle is acute. 

But if the given angle be acute, and the side opposite to the 
given angle less than the side opposite to the required angle, 
then we have no means of ascertaining the species of the re- 
quired angle, and the solution in this case is ambiguous. 


Case 5. Given the three sides, a, b, c, required the three 
angles A, B, C. 
By formula (¢) chap. III. 


gh rec daenai 
Se be V's (s—a) (s—b) (s—c) 
D 

sin, B= CoV (s—a) (s—b) s—c) 


SapeR Taaa  apent 
aoe ea aheS (s—a) (s—b) (s—cj 


Adapting these expressions to computation by the tables, and 
taking the logs. 


log. sin. A 
=log. R+-log. 2+4 $ log. s+log. (s—a)+log. (s—b)+log. (s—c)$ —log b—log. ¢ 
log. sin. B 
=log. R+-log. 24-4 ; log. s+log. (s—a)-+-log. (s—b)+log. (s—c) }—log. a—log. c 
log. sin. C 


=log. R+log. 94+-4$ log. slog. (s—a)+-log. (s—b)+-log. (s—c) $—log. a—log. b 

Whence the three angles are known. 

The three angles may also be obtained from any of the 
groups of formulee (¢), (¢), (m), in chap. III. 

It is manifest, from the remarks made at the conclusion of 
the last chapter, that, when one or more of the required angles 
is very small, the group (¢) may be used with the greatest ad- 
vantage, and when one or more of the angles is nearly 90°, 
we ought to employ the group (%.) The group (») may be 
made use of in any case. 


ANALYTICAL PLANE TRIGONOMETRY. 103 


Case 6. Given the three angles A, B, C, required the three 
sides a, b, c. 

It is manifest that this case does not admit of solution, for 
any number of unequal similar triangles may be constructed, 
having their angles equal to the angles A, B, C. 

We shall conclude this chapter by giving some numeri- 
cal examples. 


Example 1. Given A = 68° 2’ 24”, B = 57° 53’ 16.8 
a = 3754 feet, required C, 8, c. 
Then by case 2. : 
C = 180° — (A+B) 
, = 180° — 125° 55/ 40’.8 
= 54° 4’ 19.2 
, sin. B 
~ + sin A 
log. b = log. a + log. sin. B—log. sin. A 
Now log. a = 3.5744943 
log. sin. B = 9.928888 


13.5023831 
log. sin. A = 9.9672882 


log. 6b = 3.5350949 = log. 3428.43 


9 


ef b = 3428.43 
Similarly, 
log. c = log. a + log. sin. C—log. sin. A” 
log. a = 3.5744943 
log. sin. C = 9.9083536 
13.4828479 
log. sin. A = 9.9672882 
log. c 3.5155597 = log. 327.628 


c 3277.628 feet. 


Example 2. Given a = 145,b = 178.3, A = 41° 10’, re- 
quired B, C 
. This example belongs to case 4, and since the given angle 
A is acute, and the side b opposite to the required angle B 
greater than the side a, the solution will be ambiguous. 
We have log. sin. B = log. sin. A + log. b— log. a 
log. sin. A = 9.8183919 
log. b = 2.2511513 


12.06954382 
log. a = 2.1613680 


log. sin. B = 9,9081752 


* 


“= 


3 


104 ANALYTICAL PLANE TRIGONOMETRY. 


The angle in the tables corresponding to this logarithm, is 
54° 2/ 22’, but we cannot determine a priori whether the an- 
gle sought be this angle, or its supplement 125° 57’ 38”. 

a Spy Le 

Or Bie 4195" 7 os 

If we take the Ist value, 


‘C=84° 47’ 38” and the triangle required is ABC 


If we take the second value, see last fig. 
C=12° 52’ 22’ and the triangle required is ADC 

Example 3. Given a = 178.8, b = 145, A = 41° 10’, re- 
quired B. ; 

This example also belongs to case 4, but since the given 
angle A is acute, and the side 6 opposite the required angle B 
less than the side a, it follows that the angle B must be an 
acute angle, and the solution will not be ambiguous. 

We have log. sin. B = log. sin. A + log. b — log. a 


But log. sin. A = 9,.8183919 
log. b = 2.1613680 
11.9797599 


log. a = 2.2511513 

: log. sin. B = 9.7286086 
The angle in the tables corresponding to this logarithm is 
32° 21' 54’, and since, in the present instance, the supplement 
of 32° 21’ 54” cannot belong to the case proposed, the solution 


is not ambiguous. | 
Example 4. Given a =374, b = 3277.628, and the included 


angle 57° 53’ 16”.8: required A, O, b. 
By case 3 we have 


A —B C 
10g. (tan. log. (a—b)+log. cot. @ — log. (a+b) 
a— b= 476.372, .. log. (a—b) = 2.6779444 
C 
log. cot. g =10.2572497 


12.9351941 
a+b = 7031.628, log. (a+b) = 3.8470543 
A. 
log. tan. ae 9.0881398 
Az tR 
Whence Heer Des” A 
And since A+B = 122° 6’ 43.2 


And A—B—= 13° 584.8 
2A — 136° 4’ 48” 
2B = 108° 8’ 38".4 
A = 68° 2' 24”, B = 54° 4’ 19.2 


° 
ANALYTICAL PLANE TRIGONOMETRY. 105 


The angles A and B being determined, the side ¢ may be 
readily found from the equation. 
c sin. C . 
a sin, A oro 
log. c = log. a + logesin’ C — log. sin. A 


Example 5. Given g==83, b—42.6, c—53.6, required A, B, C. 
Taking the formula marked (<) in chap. IIL we have 
log. sin. A 

=log. R+-log. 244} log. slog. (s—a)-Hlog.(s—b)--log.(s—c) t —flog.b-+log.c? 
log. sin. B . 

=log.R+log.2+4 ; log. s+-log. (s—a)+-log.(s—b)-+log.(s—c) re 5 log.a+log.c$ 
log. sin. C ; 

=log. R++log. 2-++-4 ; log. s+-log. (s—a)-+-log.(s-—b) Llog.(s—c) } — 5 log.at+log.b : 


Now log. R=10. 
log.2=0.3010300 
a=33 ..log.a =1.5185139 log.b+log.c=3.3585744 
b=42.6..loz.b =1.6294096 « log.a+log.c=3.2476787 
c=53.6..log.c =1.7291648 log.a+log.b=3.1479235 
s=64.6 ..log.s =1,8102325 
s—a= 31.6 .*. log.s—a=1.4996871 
s—b=22 .. log.s——b=1.3424227 
S-c=]1_ .*. log.s—c=1.0413927 
.. log.s+log.(s—a) +log.(s—b) +log.(s—c)=5.6937350 
And , 


% jlog.s+log.(s—a) +log.(s—b) +log.(s—c)} = 2.8468675 
. log. R+log.2+4 flog.s+log.(s—a)+log.(s—b) 
+log.(s—c)} - - - - -=13.1478975 


Subtracting from this number the values log. b+log. ¢; 
log. at+log. c; log. a+log. b;in succession we find, 


log. sin A. = 9.789323] ..-: A =987° 59/53! 
log. sin. B = 9.9002188 .. B = 52° 87’ 46"55 
log. sin. C = 9,.9999740 .-. = 89° 22’ 20/55 


Having determined A and B by the above method, we find 
the above accurate value of C, by subtracting the sum of A 
and B from 180°. If, however, it had been required to deter- 
mine C alone (being an angle nearly equal to 90°) we could 
not have found its value with sufficient accuracy from the com- 
mon tables, for it will be seen, upon referring to them, that 
the number 9.9999740 may be the logarithm of the sine of any 
angle from 89° 22” 20” up to 89° 22’ 25” consequently the 
above method cannot be applied with propriety to determine 
the exact value of C, unless we previously determine A and B. 

10 


< 


* 


106 ANALYTICAL PLANE TRIGONOMETRY. 


The angle C may however be determined directly, and with 
great accuracy, from any of the three formule (¢), (%), (1), in 
Chap. III. 

Let us take these in succession, (¢). 


C re 
log. cos.5-=log. R+4 flog. s+log. (s—c)}—4(log. a+ log. 6) 


log. s=1.8102325 ) .*. 4 flog.s +log.(s—c) } = 1.4258126 
log.(s—c)=1.0413927 log. R=10 


2.8516252 11.4258126 
log.at+log.b=3.1479235 .. 3 (log. a + log. b) = 1.5739617 


e 
log. COS5= 9,8518509 


© © 


=44° 41' 10” .8, 
C=89° 22’ 20” 12 
By (). 
log.sin.5 =log.R+3 flog. (c—a) +log.(s—b)?—1 Slog.a+log.d} 
log.(s—a) = 1.4996871 } .*. +$log.(s—a) Haye t= 1.4210549 


log. (s—a@) =1.8424227 og. R=10. 
2.8421098 11.4210640 


log.a--log.b=3.1479235 + flog.at+log. b}= 1.5739617 
LG 
log. sins = 9°8470932 
==-44° 4]’ 10”; 
C=89° 22’ 2018 


wlLQ ss 


By (n). 
log.tans =log.R+2§log.(s—a)+log.(s—d)}-3§log.s+log.(s—c) 


log.(s—a) =1.4996871 
log.(c—a) =1.8424227 } 2 sie (eee) leg: ew) ¥ 1.4210549 
2.8421098 og. =10. 


log.s=1.8102325 11.4210549 
log.s—-c=1.0413927  .. 4 flog.tlog. (s—c) =1.4258126 


x 


2.8516252 C 
log. tan.5 =9.9952423 


C 
——=44° 41! 10,8, 


2 
C=89° 22’ 1012 


ANALYTICAL PLANE TRIGONOMETRY. 107 


CHAPTER VIII. 
ON THE USE OF SUBSIDIARY ANGLES. 


_ Subsidiary Angles are angles which, although not immediately 
connected with a given problem, are introduced by the com- 
putist in order to simplify his calculations. Their use, and the 
method in which they are employed, will be understood from 
what follows. 

When the two sides of a triangle, and the included angle, 
are given, according to the method pursued in the last chap- 
ter, we must determine the two remaining angles before we 
can compute the third side. It frequently happens, however, 
in practice, that the side only is required, and it therefore be- 
comes desirable to have some direct method of computing the 
side independently of the two angles. 

Suppose that a, b,C are given, andc is required. By chap. 
Ill. prop. 4, 

c’=a'?+b?—2ad cos. C. 
the side c is determined theoretically at once by this expression, 
but the formula is not adapted to logarithmic computation, and 
would, if employed practically, lead to a very tedious and 
complicated calculation. We can, however, put this expres- 
sion under a form adapted to logarithmic calculation, by hav- 
ing recourse to an algebraical artifice, and introducing a sub- 
sidiary angle. 
c=a'+)°—2 ab cos. C 
Adding and substracting 2 ad on the right hand side. 
Ce=e?+h—2 ab+2 ab—2 abcos. C 
=(a—b)’+2 ab (1— cos. C) 


C 
=(a—b)’+2 ab+2 sin.'5 


4 qb ane | 
2 


EUR OMe as [aes 
(a—b) aay 


*. _& 
4. ab sin. P) 


Assume = tan.’ o 


(a—y 
c* =(a—b)? (1+tan.? 9) 
=(a—b)’ sec.” 
¢ =(a—) sec. o 
<p log. ¢ =log. (a—b) +log. sec. p—log. R 
he angle ¢ is known from the equation. 


108 ANALYTICAL PLANE TRIGONOMETRY. 


2/ab. sin. u 
man. =", 2 


(a — b) 
Whence, 


log.tan.p=log. 2+ 1 (log.atlog.b) + log. sin. —log.(a—2) 


» being thus determined, log. sec. p can be found from the 
tables, and the value of c becomes known. 

The angle 9, which is introduced into the above calculation, 
in order to render the expression convenient for logarithmic 
‘computation, is called a subsidiary angle. 

The above transformation may be effected in a manner 
somewhat different, as before. 

C?=a?+bh—2 ab cos. C 
=a’?+0?+2 al—2 ab—2 ab cos. C 
= (a+b)*—2 ab (1+-cos. C) 


=(a+b)—2 abxX2 cos.’ = 


: 2 
4 ab cos.’ 4, 
=(a—b)’ (I— 2 
(aFby 
4 ab cos.’ Cc 
Assume | 2=sin.’*9 
(a+b) 


c’=(a+b)* (1—sin.? ¢) 
=(a+b)* cos.’ @ 
c=(a+b) cos. 
{t log. c=(s+b)+log. cos. p—log. R 
As before the angle p must be determined from the equation. 


2./ab ; aoe 
sin.o= 2 
(a + 4) 
In order to prove that we can always assume 
— C 
2./ab . cos.— 
(a+) 
2./ab cos.2 
we must show that 2 is always less than unity, or, 
(a+c) 


in other words, that 2./aé is always less than (a+0), this iS 
easily done. 


a. SIND 


~ 


ANALYTICAL PLANE TRIGONOMETRY. 109 


If a+b > 2Yab 

Then a?+2ab+0’> 4ab 
a’ +0 > 2ab 
a+b?—2ab> 0 

Or (a—b)? >O0 


But since (a—b)’ is necessarily a positive quantity, it must 
always be greater than 0 (except in the particular case a=, 


where it is=0), and therefore?V@ ©S-3 js always less than 
(a+6) 

unity, and consequently an angle may always be found whose 

sine is equal to it. 


In solving the same case of oblique-angled triangles, we de- 
termined the difference of the angles A, B from the equation. 


a A—B _ a—pb es. 
ee Ce eee eo 


Whence peta — =log.(a—!) +log. cots —log. (a + d) 


In the solution of certain astronomical problems, the loga- 
rithms of the sides a,b are given, but not the sides them- 
selves, and these logarithms being given, we can very easily 


2 


calculate without knowing the sides. 


A—B _ a—b C 


b 
Assume =tan. 9 
A—B_ 1—tan. ( C 


tan.— 3 F Titan ys cot. ey 
C 
=tan. (45°—2) cots 
a. C 
“. log. tan.—> =log.tan. (45°—¢) +log.cot.5 —log. R 


10* 


: “s 


5 MS eee j 
Le og 
< 3/4 ‘ 


110 ANALYTICAL PLANE TRIGONOMETRY. 


The angle 9 is known from the equation. 
tan. p= — 
a 


Whence log. tan. p=log. R+log. b—log. a 


—— 


The angle — thus becomes known from the logs. of a 


and b, without calculating a and b. In the same way we have 


C 
cot. =tan. (45°+) tan. oe 


And .°. log. cot. 


=log. tan.(45°+ 0) +log. tan —log.R. 


CHAPTER IX. 
ON THE SOLUTION OF GEOMETRICAL PROBLEMS BY TRIGONOMETRY. 


A great variety of geometrical problems may be solved 
with much elegance by the introduction of geometrical for- 
mule. We shall give a few examples. 


PROBLEM I. 


To express the area of a plane triangle in terms of the 
sides of the triangle 


Let CD be a perpendicular from C upon AB 


Area of a triangle ABO 5 
Cc . 
ae AC sin. A th 
1 eel 
Oe sin. A - 
_ be 2 Vs(s—a)(s—b)(s—c)..Ch. IIL. 
meee? SDE 


= Vs (s—a) (s—b) (s—c) 


ae 
. 


ANALYTICAL PLANE TRIGONOMETRY 11i 
PROBLEM II. 


To express the radius of a circle inscribed in a given triangle, 
in terms of the sides of the triangle. 
* 


Let the radius required be called 7 
b 
Area of AOC=—- 


GS. 
re Ps Ps 
AOB=-~- 
ra 
COB=-— nN P, , 
. Whole area of triangle ABC=5 (a+b+c)= 
i.e. Ws (s—a) Shard) (s—c)=r.s __ by last problem 
r= ss (s— s (s—a) (s- (s—b) —b) (s—c) 
s 


PROBLEM ilt. 


To express the radius of a circle circumscribed about a given 
triangle, in terms of the sides of the triangle 
| 


Let fall CD perpendicular on AB 
Let the radius be called R. 
By Prop. XXXIX, B. IV. Hl. Geom. 
CQ. CD=—AC.CB 
CO Cr enn OEP eA, 
» 2 RX2 area=abe 


abc 


abc 


45 (a (s— -a) (s—b)(s—c) 


112 ANALYTICAL PLANE TRIGONOMETRY. 


PROBLEM IV. 


Given the three angles of a plane triangle, and the radius of 
the inscribed triangle, to find the sides of the triangle. 
- 


Let A, B, C, be the three given angles, 7 the 
radius, 


AB or c=AP,+P,B 


A =) 
=F) COL. Tp Phe = 
‘sin a) 
a! REE GeMAMae ORAS 
sinve+ SIN.) 
2 2 
et sin. (3 + =) 
“ ACO ATs a a 
sin 53 sin 


2 
‘ (5 “ =) 
sin. {— — 
BO or. a=. ite 2 


PROBLEM V. 


Given the three angles of a plane triangle, and the radius of 
the circumscribing circle, to find the sides of the triangle. 


As in Problem III. 
€Q..CD=—AC..CB 
CQ. CB sin. B=AC . CB 
oe AC=2 R sin. B 
So, BC=2 Rsin. A 
AB=2 R sin. C 


ANALYTICAL PLANE TRIGONOMETRY. 113 


CHAPTER X. 


PROBLEMS IN TRIGONOMETRICAL SURVEYING. 


THE DETERMINATION OF ‘TOPOGRAPHICAL DATA BY GEOMETRICAL 
CONSTRUCTION AND TRIGONOMETRICAL ANALYSIS. 


PROBLEM I. 


To determine the height of an inaccessible object. 


Let AB be the object, and in a straight line 4 
towards it measure any distance DC, and ob- 
serve the angles of elevation ADB, ACB at 
the stations D,C. PutCD=h, ACB=a, ADB 
=b: then DAC=a—d; hence we have 

AB | 4 AC | sin. 6 
AG en Da sind) 
and, multiplying these two equations, we have 
AB sin. a sin. b 
CD “sin. (a— 0) 
. log AB=log.h+log.sin.a+log.sin.b+log.cosec.(a—b)-80 (1) 


Cor. Since DB=AB cot. b, and CB=AB cot. a; therefore, 


h 
by subtraction, CD=AB(cot.b-cot.a,)or AB= cnib BORE Ae 


fz. Let DC=h=200, BDA=d=81°, BCA=a=46°; to 
find AB and CB. 


, or, AB=A sin. a sin. 6 cosec. (a — 2) 


log. h- - - = log. 200 = 2.8010300 
log.sm.a- - = log. sin, 46°= 9.8569341 
log.sin.b- - = log. sin. 31°= 9.7118393 


log. cosec.(a—b) =log. cosec. 15°=10.5870038 


log. AB - - =Ilog. 286.29 = 2.4568072 ... AB=286.29 
Also, BC=AB cot. a, .. log. BC=log. AB+log. cot. a— 10. 


&. 


114 ANALYTICAL PLANE TRIGONOMETRY. 
PROBLEM II. 


To determine the height af an inaccessible object, which has no 
level ground before it. 


Let AB be the object, and C, D, two 
stations in a vertical plane passing through 
AB; measure the distance CD, and atC p 
take the angles of elevation or depression 
of the station D, and the top and bottom | : 
of the object. Also at D take the eleva- H Cc K 
of the top of AB. 

Put CD=h, DCH=a, BCK=s, ACK= re ADG= d’; fihen 
ACB=c— 6, ADC=a+d, and CAD=c—d 

N AB _ sin. ACB sin. (c—) _ sin. (c—t)' 
OW) AC sin. ABC’ sin. ABK cos. } 
AC sin, ADC _ sin. (a+d) h ihc 
CD sin. CAD sin. (c—d)* Pg Qua 
equations, 


: AB _ sin. (c—8) sin. (a+d) 
ge DONS 1) aos Bibs (Cea) wo 


AB=Asin. (c — 5b) sin. (@+d) sec. b casec. (c—d) 
.“. log. AB=log. h+log. sin. co oe) a Oe sin. bee aD he 
sec. b--log. cosec. (c — d)—40 


Cor. When a=90°, and 6=0°, then we have 
log AB=log.A+log.sin.c+log.cos.d+log.cosec.(c—d)—30 (2) 
Hz 1. Let h=18 feet; c=40°; d=37°30'", a= 205 * and b=0°; 
to find AB. 
loge Revers. ; «f= log.-18 0) fun. 12552725 
log. sin.c. . =log. sin. 40° . = 9.8080675 
log. cos.d. . =log. cos. 37°30’= 9.8994667 
log. cosec.(c—d) =log. cosec.2° 30’=11.8603204 
log. AB . . =log. 210.4894 = 2.3231271 
.AB=210.4394. 
Hx. 2. The angle of elevation of the top of a tower, stand- 
ing on a hill, was 33° 45’, and measuring on level ground 300 
yards directly towards the tower, the angles of elevation of 
the top and bottom of the tower were 51° and 40° respectively. 
What is the height of the tower ? Ans. 140 yds. 


; and, therefore, 


Remark.—When the station D is higher than A, the top of 
the tower, then the angle d must be considered negative, and 
therefore we should have 

AB= h sin. (c—b)sin. (a—d) sec. b cosec. (c+d) 


®, 


ANALYTICAL PLANE TRIGONOMETRY. ¥15 


PROPOSITION I. LEMMA. 


[f straight lines be drawn from any point, either within, or 
out of, a polygon, to all the angular points, the continued pro- 
ducts of the sines of the alternate angles, made by the sides of 
the polygon, and the lines so drawn, will be equal. 


Let angle CDP=f, and PEA=i; 
sin. b PB sin. d 


then PB re sin. a@ PC fF sin. c 
PC _ sin. f PD __ sinsit 
PD sin. é PE sin. £ 
Pee Sik 
PA sin. 2. 


BotebaA .PB.?C. PD.PE=PB.PC: PD. PE ..RBA; 
that is, the product of the numerators= product of the denomi- 
nators, in the first members of these equations ; hence, this 
being true in the second members also, sin. b sin. d sin. f sin. 
h sin. K=sin. @ sin. c sin. e sin. g sin. 2. 


PROBLEM III. 


Given AB, and the angles a, b, c, d, to find © D 
Ly and thence CD. S403 5-0 
Put BCD+ ADC=)+c=2s . 


BCD—ADC=.:.. 2z 2 
Then BCD = s+z, ADC = s—z; also, sin. 


ADB= sin. (b+c+d), and sin. ACB = sin. = 
~ (a+b+e); hence, by Lemma, (I) we have 
sin. asin ¢ sin. (b+c+d) sin. (s+z)=sin. 6 sin. d sin. (a+b+c) 
sin. (s--%), or sin. @ sin. csin. (b+c+d) {sin. s cos. x+cos. s 
sin. z{ = sin. 0 sin. d sin. (a+b+c) {sin. s cos. z—cos. $ 
sin. 2}. 

Thon, dividing by cos. s cos. z, we have 


2 
es 


<y 


116 ANALYTICAL PLANE’ TRIGONOMETRY. 


sin. @ sin. ¢ sin. (b+c+d) (tan. s+tan. z)=sin. bsin. d sin. 
(a+b-+c) (tan. s—tan. x), 
: __sin.d sin. d sin.(a-+b-+c)-sin.a sin.c sin.(6+ce+d) 
“tan. 2= nO sin.d sin.(a+6+c)+sin.a sin.c sin.(6-+ce+d) 
Dividing numerator and denominator by sin. @ sin. ¢ sin. 
sin. d sin. d sin. (a+b+c) 
b ng ——_—_=—_ 
ee 2) + Pod RaEAS sin. a sin. c sin. (6+-c+d) 
l= tan.'45°; then 
tan. 8 — tan. 45° __ sin. (8 — 45°) 
tan. 8 + tan. 45°“ * — sin. (8 + 45°) 


hence x, s-+2, s—zx are all known, and thence CD is known. 
For CD sin. d BD sin. 6 : 


BD sinters) TAB 2 ae oes re 
CD=AB sin. 6 sin. d cosec. (s-++x) cosec. (b-+-c-+d). 


tan.s. 


=— tan. & and 


tana = 


Cor. When CD is given, and the same angles, to find AB, we 
have 
AB=CD sin. (6+c+d) sin. (s+2) cosec. 5 cosec. d. 


EXAMPLE. 


Given AB=600 yards, a2=37°, b=58° 20’, c=538° 30’, d= 
45° 15’, to find CD. 
Here, tan. 8=cosec. a sin. 6 cosec. ¢ sin. d sin. (a+b-+c) 
cosec. (h-+c+-d), 
log. cosec. a 
ee WORT COSCO MOT © nt es = 10.2205370 
log. sin. b 
= log. sin. 58° 200° 5. = 9:929989T ~9.9299891 
log. cosec. ¢ : 
= log. cosec. 53430 «..= 10.0948213 
log. sin. d 
= log. sin. 45. .15...= 985138717 9.8513717 
log. sin. (a+b+c) 
== Jog. sin. 148 50...=. 9.7139349 
log. cosec. (b+c-+d) 
= log. cosec. 157 5 ...= 10.4096181 10.4096131 


od 


log. tan. 6 

= log. tan. 58° 56’ 39” = 10.2202671 ne 
log..sin. (8 — 45°) | 

= log. sin. 13 56 39 = 9.8819742 


log. cosec. (8+45°) 
= log. cosec. 103 56 39 = 10.0129906 


ia 
*'e 
ANALYTICAL PLANE TRIGONOMETRY. 117 


log. tan. s 
= log. tan. 55 55...== 10.1696508 
log. tan. x 
= log. tan. 20 9 8 = 9.5646156 
log. cosec. (s+2) 
= log.cosec. 76 4 3 =....-..- 10.0129687 
log. AB. 
= log. tT ag ere os ae 2.7781513 
* CD A~4)= 59.608 Fee ees ren 2.9820939 


PROBLEM IV. 


The distance of two objects at B being known, I find, by ob- 
servation, the angles ACD ADB BDC, and BCA, taken at the 
stations D, Cyrequired the distances DOC, AD, BD, BC, and 


CA, both by construction and calculation. 


Assume dc at pleasure, 4 B . : 
and make the angles adb, 
bdc and bca, respectively >< 
equal to the angles ADB d 


BDC and BCA; join a,b, 
and abcd will be similar 5 | 
to ABCD, and if ab re- y 

present the side AB, then will de repersent DC. &c. 


By analysis, 


In the triangle adc, all the angles and the assumed side, dc 
are given to find ad and ac. Then in the triangle bcd, all the 
angles and the assumed side dc, are given to find be, bd. 
Lastly, in the triangle adb, we have the sides ad, db, and the 
angle adb, to find the side ab. Then we have, 


ab: AB:: de: DC:: ad: AD:: bc: BO :: ac: AC:: bd 
: BD, whence we have the distances, DC, AD, BC, AC and 
BD. 


Ml 


118 ANALYTICAL PLANE TRIGONOMETRY. 


PROBLEM V. 


-_ Given AB, a, b, and the angles c, d, taken at some point Pin 
the same plane ABC, to find x; and thence PA, PB, PC. 


Put PAC+PBC=180°—(a+b+e+d)=2s A B 
PAC] BC =.ch-, ie duets Hen Qe; 
Then, PAC=s+z, PBC=s—z, and, by 
lemma 1, sin. @ sin. c sin (s—z)=sin. b sin. d 
sin. (s+<) 
_sin,dsin.d sin. (s—) _ tan. s—tan. x 
"*sin.asin.c sin. (s+x) ~ tan. s+tan. 2 


sin. 6 sin. d P 
Put tan 8.‘ >... cosec a4 Sin 6 
sin. a sin c 


cosec. c sin. d; then we have 
tan.s— tan. 2 _ tan. 2 I1—tan.f8 tan. 45°—tan. 8 


tan. s-+tan. =tan. 6 .-. tan. s_1+tan. 8 ~ tan. 45°+tan. 8. 
a ew (am 45°—tan. 8 tng sin. (45 mie i 
2" F tan. 45°+tan. 6 ‘ sin. (45°+8) °°" 
Hence z is known, and thence s+z and s—z2 are known. 
PC sin. (s+xz) AC sin. b 


PC = AB cosec. (a+) sin. 4 cosec. c sin. (s+2). 


PROBLEM VI. 


When the points P and C are on opposite sides of AB. 


Put PAB+PBA=180°—(c+d)=2s 
PAB—PBAS®... TUS». G22Qz ; 
then, PAB=s+2, PBA = s—2x; and, bya 
lemma 1, sin. a sin. c sin. (s —z) = sin. 6} 
sin. d sin. (s+2) ; 
hence, as in the last problem, we have 
sin. (45°—/) 
tan? c= sin. (45°-+8) tan. S$; 
where tan. 8 = cosec. a sin. 6 cosec. c sin. d; and 2s 
= 180°—(c+d). 


ANALYTICAL PLANE TRIGONOMETRY. 119 


PART II. 
Many curious and highly useful problems in trigonometrical 
surveying, may be elegantly solved through the properties of 
the circle, by geometrical construction, and by analyzing this 
construction by trigonometrical analysis; a few of which we 
will give in the continuation of this chapter. 


PROPOSITION II. LEMMA. 


If two points be assumed in the circumference of a circle, they 
will subtend the same angle from any point whatever of the 
circumference on the same side of the chord joining these two 
points, which will be half the angle at the centre when the 
centre is on the same side of that chord as the point of obser- 
vation, and will be equal to half its complement to 360° when 
on the opposite side. 


This proposition is evident from Prop. XIX. Cors. 1 and 8, 
B. Ill. El. Geom. 

Hence, if F,G are two points assumed 
in the circumference ABFG, then will 
_F, G appear under the same angle from 
any points A and B situated in the cir- g 
cumference, and on the same side of the 
chord FG; which will be half the angle 
C at the centre; and the points F’,G will 
also appear under equal angles at every 
point D, E on the side DE of the chord 
FG, which will be equal to the angle mea- 
sured by half the arc FBAG equal the complement of the angle 
FCG as enunciated. 


Cor. 1. Hence, any two objects in the circumference of a 
circle will always appear under the same angle, in any point 
of the arc of either segment, and in no other point situated out 
of that circumference, on the same side of the objects, will the 
angle be the same. 


Cor. 2. If the angle under which any two objects appear 
be less than 90 degrees, the place of observation will be some 
where in the arc of the greater segment; and if the angle be 
greater than 90 degrees, the place of observation must be some 
where in the arc of a segment less than a semicircle, and the 
angles under which the objects appear, will be the same in 
any point whatever of those arcs. 


120 ANALYTICAL PLANE TRIGONOMETRY. 


Scholium. Hence, having the angles subtended by any two 
objects from any two given positions, not all in the same cir- 
cumference, the positions of the objects may be determined by 
the intersections of the circumferences of two circles, each of 
which is so described as to pass through the two objects and 
one of the given positions. 


PROBLEM VII. 


Three points in the same plane being given in position, to 
determine the position of any other point or place of observa- 
tion in reference to the given points. 


This problem admits of six cases. 

The three given points may be the vertices of a triangle, 
and the required point, or station, may be without the triangle, 
and opposite one of its sides; it may fall in the same right 
line with two of the given points, it may fall directly between 
two of them, it may fall within the triangle or it may fall 
without the triangle but opposite one of the angles ; and lastly, 
the given points may be all in the same straight line. 


Case 1. 

When the given points are the vertices of a given triangle, and 
the station regarded, falls without the triangle and opposite 
one of its sides. 


Let A, B, C, be the given points C 
whose positions in reference to each B 
other are known, and let S be the 
point required. Having taken, the 
angles ASC, ASB, describe on AC the 
segment of a circle that shall contain 
an angle equal the observed angle ASC; 
and on CB describe a segment that 
shall contain an angle equal to the angle CSB, and the point 
of intersection of the arcs of those segments will determine 
the position S. | 

Or, 

Make the angle EBA = the observed E 
angle ASC, and the angle BAE = the 
angle BSE; through A, B, and the in- EZ e 
tersection at KE, describe the circle A F B 
AEBS; through E and C draw EC, 
which produce to meet the circumfer- 
ence atS. Join AS, BS, and the dis- 
tances AS, CS, BS will be the required 
distances of the station S, from the S 


points A, B, and C. 


S 


“ANALYTICAL PLANE TRIGONOMETRY. 121 


By trigonometrical analysis. 
In the triangle ABC, the three sides are given to find the 
-angle BAC. And in the triangle AEB, we have angle EAB 
= angle BSE, angle ABE = angle ASE, and therefore the 
angle AEB, with the side AB, to find AE and BE. Also in 
- the triangle AEC, we have the sides AC, AF, and the included 
angle to find the angle AEC. _ Whence the sum of the angles 
AES, and the observed angles ASE, subtracted from 180° 
gives the angle SAE. Then in the triangle AES, we have 
all the angles and the side AE, to find the side AS. And the 
angle ACS = 180°— angle SAC — angle ASC; then in the 
triangle ACS we have all the angles and the side AC to find 
CS. Angle AEB—angle AEC = angle BES, whence we 
have the side BE of the triangle BES, and the angles E and 
S, to find BS; then AS, CS, and BS are the station distances 
required. 


Scholium. Ist. If the angle BSC, be less than CAB, the 
point E will be below the point C. 

2. When the points E and C fall so near each other that the 
production of EC toward S is attended with uncertainty, the 
former method of construction is preferred. 


Case 2. 

Let it be required to determine the position of an observer at S 
in reference to the three objects ABC, when SAC are in the 
same right line. 


Having taken the angle at 8, and cal- Ader C 


culated the angle CAB from the sides of § 
the triangle ABC which are known by 
hypothesis, we have the angle ABS=ang. 
CAB—ang. ASB. Then at B with the 


side BA, construct the angle ABS. produce the side, BS till 
meets the production of CA in 8, and SA, SC, SB will be 
the several distances of S from the points A, C, B, whence hav- 
ing the angles 8, C and B and the side BC, the sides SB, SC 
and SA may be obtained. 
By trigonometrical analysis. 

Angle SAB = 180°—BAC, angle SBA = CAB—ASB. 
Hence in the triangle SAB, we have all the angles and the 
side AB to find the distances AS, BS, &c. 


Case 3. 

~To determine the position of S in reference to three given ob- 
jects ABC, where the required point is directly between A 
and B. 


B 
it 


1]* 


aad + 
: » 


122 ANALYTICAL PLANE TRIGONOMETRY. 


Having constructed the triangle ABC 
which is given by hypothesis, and hav- 
ing observed the angle BSC, construct 
from any point A on the line AB an angle 
BAE=the observed angle and draw CS 
parallel to EA, and S is the required sta- 
tion. rs 


By Analysis. 

In the triangle ABC all the sides being known, let the an- 
gle A,B be obtained. Then in the triangle BCS having the 
angle S and B, and the side CB we may proceed to find the 
distances BS, CS. Then we shall find AS=AB—SB. 


Case 4. 


To determine the position of a station S in reference to three 
given objects when the required station falls within the tri- 
angle formed by connecting those given objects. 


Let the given objects be three vs 
towns A,B and C which are all 
visible from a station S, which is 
included in the triangle formed 
by lines drawn from A to B, from 
B to C, and from C to A. 5 


A B 


First take the angles ASB, BSC, CSA, then on‘either side 
AC describe an arc ASC, which shall contain the observed 
angle ASC, and one either of the other sides AB, describe 
an arc which will contain the angle ASB, and the point of in- 
tersection of those arcs is the station.S, all of which is evident 
from Lemma II. 


Otherwise, on AB make an ¢ B 
angle ABl=the supplement of 
the angle ASC; and make an 
angle BAE equal to the supple- 
ment of the angle BSC, then will 
ABE=—ASE and BAE = BSE, , 
since ASE and BSE are respec- 
tively the supplements of the an- 
gles ASC and BSC, through the 
points A, B, and E describe a 
circle, join EC cutting the circum- 
ference in the point S which is the station required. 


ate 


» * 


ANALYTICAL PLANE TRIGONOMETRY. 123 


By Analysis. 

Ist. In the triangle AEB the angles Band A being the sup- 
plements of the observed angles CSA, CSB, are therefore 
known, and consequently the angle E and the side AB, to find 
the sides AK, BE. 

2d. In the triangle ECA we have the sides EA, AC and 
their included angle EAC=BAE+CAB, to find the angle AEC. 

3d. And the triangle CEB the sides CB, BE, and the in- 
cluded angle CBE are given to find the angle CEB. 

4th. Therefore in the triangle SAB the angle A, being 
equal the angle E, since they are angles in the same segment, 
is also known and also the angle ASB, hence we have all the 
angles and the side AB to find the sides AS, BS which are 
two ofthe station distances required. 

5th. If from the angle CAB we take the angle SAB, we 
shall have the angle CAS. Therefore in the triangle ACS 
we have all the angles with the sides AC and AS to find the 
other station distance CS. 


Case 5. 


Tet it be required to find the distance of any station S from each 
of three objects A, B, C, when one of the angles C of the tri- 
angle formed by connecting the three objects falls toward the 
station NS. 


Make the angle DAB=the observed 
angle CSB and the angle DBA, equal to 
the observed angle CSA. On AB de- 
scribe a circle that shall contain in its 
greater segment the observed angle ASB, 
through D and C draw the line DC, till it 
intersects the circle at S, which intersec- 
tion determines the position S. 


By Analysis. 

Ist. In the triangle DAB, all the angles and side AB are 
known to find AD, DB. 

2d. In the triangle ADC are given the sides AD, AC and 
then included angle, to find the angle ACD. 

3d. The angle CAB=ACD—ASD, since ACD, the out- 
ward angle is equal to the sum of two inward opposite angles 
CSA, CAS. Therefore in the triangle ACS we have all the 
angles and the side AC to find the distances SA, SC. 

Lastly, in the triangle BSC the sides CB CS, and the angle 
BSC are given to find the distance CB. 


124 ANALYTICAL PLANE TRIGONOMETRY. 


Case 6. 
Lo determine the position of any station Sin reference to three 
objects ABC all in the same straight line. 


On AB describe an arc of a seg- 4B ‘ 
ment to contain an angle equal 
to the angle ASB, and on BC de- 
scribe an arc containing an angle 
CSB= the observed angle, subtend- 
ed by BC, and the point of inter- 
section of those arcs will determine 
the position of S ; whence if we S 
draw the lines SA, SB, SC, those lines will the several dis- 
tances of S from the objects A, B, C. 


For (Lemma II) the points A, B appear under the same angle 
in every point in the arc BSA, and in no point out of that arc, 
and B,C, appears under the same angle in every part of the 
arc CSB and no point out of the arc. Hence the point of in- 
tersection of those arcs is the only point where both of those 
conditions are united, or where both of the objects appear 
under the observed angles. 


Otherwise, at A and on the line AB 
make BAK=the observed angle CSB 
and at C an angle=the observed angle 
ASB; on AC describe a circle that 
shall contain an angle=the sum of the 
observed angles at 8S or which is the 
same, describe a circle which shall pass 
through the three angles A, E and C, 
from E through the point B, drawn EBS 
to cut the circle inS, andSA,SB, SC S 
determine the relative position of the station S in reference to 
the three A, B and C. 


By Trigonometrical analysis. 

Ist. In the triangle CAE all the angles are given and the 
side AC to find AE. 

2d. In the triangle AEB the sides AE, AB, and their in- 
cluded angle are given to find the angles AEB and ABE. 

3rd. Inthe triangle BSC we have the angle CSB and 
the angle SBC=ABE, and the angle SCB=angle AEB since 
they are both angles in the same segment ACS, hence all the 
angles and the sides BC are given to find the side, SC, SB. 
and SA. 


ANALYTICAL PLANE TRIGONOMETRY. 125 


PROBLEM VIII. 


The distances of three objects A, B and C being given and con- 
sequently the angles which they form with each other. There 
are also two stations D, I, such 
that at D the objects A,C and K may | © 


be seen but not B; at E the object, B, 

C and D may te seen but not A. , B 
Hence the anglesCDEK ADC BED © 

BEC and consequently the angles 

CDA and CEB are given or known 

from observation to find the dis- 

tances DA, DC, DE, EC, and EB. i 4 


c 
ae 
b 
Draw cd at pleasure, and atd make @ | 
an angle cde=the given angle CDE | 
make also the angle dec=DEC, the 
angle ceb=CEB, and cda =CDA / 
a () 


produced ad, bc, till they meet in s, , 
and draw sc. Ne wy 


By Analysis. 

Assume any value for de; then in the triangle cde all the 
angles are given and the side de, to find the sides cd, ce, the 
angle eds=180°—-ade and the angle des=180°—ed ; hence 
we have in the triangle dse all the angles and side de, to find 
ds, andes. Inthe triangle cds the angle cds =180°—adc, hence 
we have two sides cd ds, and their included angle to find the 
angle dsc=csa and side cs, then from the angle dsc take the 
angle dsc and we have the angle cse=csb. 


C 


Then with the angles csa, 
csb and the triangle ABC, as 
data, make the following con- 
struction by case first Prop. 
VII, and find the station dis- 
tances, SA, SB, SC. Then 
we shall have 


126 ANALYTICAL PLANE TRIGONOMETRY.’ 


cs: CS::de DE:: dc: DCO::ec: EC::ds:DS:: se: SE 
from SA take SD and there remains DA, also from SB take 
SE and there remains BE, hence DA, DC, DE, EC and EB 
are found. 


Scholium. 

When AD and BE are pa- 
rallel, the foregoing method of 
solution fails. In which case, 
on AC describe a segment to 
contain an angle equal to the 
observed angle CDE, and on 
CB a segment to contain an 
vangle equal to CEB, draw 
the chord CF to cut off the segment CADF containing the an- 
gle CDE, and another chord CG cutting off a segment CBEG, 
containing the angle CED, the points F and G will be in the 
same right line with D and E, join GF which produce both 
ways till it cuts the circumference in D and E, and the points 
D and E will be the stations required. 


PROBLEM IX. 


The relation of the four points B,C,D,F, to each other are known, 
or the four sides of a quadrilateral figure and its angles are 
known, there are also two stations A, EK, such that at A only 
B, C, E, are visible and at E only the points D, F, A, so 
that the angles BAC, BAE, AED and DEF and consequently 
AEF may be known, required the distances AB, AC, ED, EF, 
EC, and AD. 


Ist. On BC describe a 
segment to contain the angle 
BAC, and draw the chord 
Cm that shall cut off an an- 
gle BCm = the supplement 
of the angle BAE. 

2nd. On DFE describe a 
segment that shall contain 
an angle DEF, and draw 
the chord Dn, that shall cut 
off an angle FDn = the sup- 
plement of the angle AEF, and the intersections m and n will 
be in the same right line with the stations A and E. 

3d. Through the points of intersection m and n, draw the 
line mn which produce to E at its intersection with the arc of 


ANALYTICAL PLANE TRIGONOMETRY. 127 


the segment and the points of intersection, A and E will be the 
points of station. , 
The analysis of this may be supplied by the student. 


PROBLEM X. 


Given the base AB, the perpendicular FD, and vertical angle 
ADB of a triangle to find the sides AD, DB. 


On the base AB describe an isosceles triangle ACB whose 
vertical angle C shall be double the given angle ADB, if that 
angle is less than 90°, but double its supplement if the angle 
ADBis greater than 90°, and from this vertice as a centre, 
with the radius CA or CB describe a D 
circle, and the vertice D of the triangle 
will be found somewhere in the circum- 
ference (LemmalII.) Ata distance FD 
equal tothe altitude of the given triangle, 
draw aright line IL parallel to AB, and 
the point where this line cuts the circum- 
ference will determine the position of 
the vertical angle ; hence the sides DA 
and DB may be drawn. 


By Analysis. 
Draw the diameter DCP, and from C draw CH perpen- 
dicular to AB. Hence having the angle ACB we have also 


the angles CAB and ABC each equal to ua Rs 


— there- 
fore in the triangle ABC having all the angles and the side AB, 
the two equal sides AC, BC, become known also. And in the 
right-angled triangle AHC, CH= VAG? AH? 

And in the similar triangles DFN, CHN we have DF: 
DN ::CH:CN 
and by composition :: DF+CH : DN+CN 
or DF+CH:DC:: DF: DN 
Also HN= /CN?__CH? ; 
and BN=i AB—HN 
FN= VDN*—DF* 
and BF=BN—NF | 
Whence we have the right-angled triangled BFD with the 
sides BF and DF including the right angle, to determine the 
third side BD which also becomes known, and consequently the 
side AD as required. 


~~ 


og Yee, ee 


128 ANALYTICAL PLANE TRIGONOMETRY. 


PROBLEM XI. 


At the distance AB from the bottom of a tower is an object 
whose length is BD, how far must I ascend the tower that 
the object may appear under any angle T'? 


Draw the line AB=the given dis- 
tance, and produce it to D, then will 
BD represent the object; draw the in- 
definite line AE perpendicular to AB, 
which will represent the side of the 
tower. Then on DB make an angle 
BCD=2T, and from the vertex Casa 
centre, describe an arc of a circle pass- 
ing through the points D and B, cutting 
the tower in F,and AF will be the dis- p 
tance required. 

For since DCB is an angle in the centre of a circle, and 
AFBvis an angle in the circumference subtending the same 
arc, hence (Prop. II.) the angle DFB=4 angle DCB=T. 


Cor. Since by continuing the are of the circle the perpendi- 
cular is cut also in G; this point also answers the condition of 
the question, for angle DGB is evidently=the angle DFB, 
hence, the question admits of two answers.* 

By Analysis. 

First, in the isosceles triangle CBD we have the side BD 
and the angle C by construction, and since the triangle is 
isosecles, the angles B and D are each = (180°—2'1')+ 2 
=90°—T. Hence, having all the angles and one side, the 
sides CB or CD are also known. 


Second, in the triangle ABC we have the sides AB and BC, 
and the angle B=CBD+2 angle BCD=90°+T, to find the 
side AC and the angle CAB, which thereby become known. 

Third, in the triangle ACF we have the sides AC and FC, 
and the angle CAF=90°— CAB, to find the side AF, the 
height required. But since the same data given to determine 
this triangle, apply also to the triangle ACG, the point may be 
also in G; hence, the problem is ambiguous both by construc- 
tion and analysis, as explained in case 4th, chap. VII. 


* This elegant construction was received from Mr. Joseph Gallup, of Norwich, 
C onnecticut, whose mathematical talent is acknowledged to be of a high order. 


~~ 


pill Pai es 
cE. ANALYTICAL PLANE TRIGONOMETRY. 129 


Scholium. If the circumference which passes through the 
points D, B should not cut the edge of the tower or perpendi- 
cular AK, but only touch it, it would admit of only one solution, 
and that point which would answer the conditions would be 
the point of contact; but if the circle should not reach the per- 
pendicular, the question would be impossible. 


EXAMPLES FOR PRACTICE. 


Ex. 1. Given the angles of elevation of any distant object, 
taken at three places on a level plane, no two of which are in 
the same vertical plane with the object; to find the height of 
the object, and its distance from either station. 

Let A, B, C, be the three stations, K the ob- 
ject, and KH perpendicular to the plane of the 
triangle ABC. 

Put BC=a, AC=), AB=c,, HAK=a, sf 
HBK=, HCK=y, and HK=z; then the 
angles AHK, BHK, CHK being right angles, 
we have AH=z cot. «, BH==z cot. 8, CH=z 
cot. y; whereby from the given data the required may be 

found. 

Ex. 2. Given a=30° 40’, 8=40° 33’, y=50° 23’; find 2, 
when the three stations are in the same straight line, AB be- 
ing=50° and BC=60 yards. Ans. 77.7175 yards. 

Ex. 3. Demonstrate that sin. 18°=cos. 72° is =} r (—1+ 
/5), and sin. 54°=cos. 36° is=1 rk (1+ /5). 

Ex. 4. Demonstrate that the sum of the sines of two arcs 
which together make 60°, is equal to the sine of an arc which 
is greater than 60°, by either of the two arcs: Ex. gr. sin. 
3’+sin. 59° 57’ =sin. 60° 30’; and thus that the tables may 
be continued by addition only. 

Ex. 5. Show the truth of the following proportion: As the 
sine of half the difference of two arcs, which together make 
60°, or 90°, respectively, is to the difference of their sines ; so 
is lto /2, or /3, respectively. 
ua Ex. 6. Demonstrate that the sum of the square of the sine 
and versed sine of an arc, is equal to the square of double the 
sine of half the arc. 

Ex. 7. Demonstrate that the sine of an arc is a mean pro- 
portional between half the radius and the versed sine of dou- 
ble the arc. 

Ex. 8. Show that the secant of an arc is equal to the sum 
of the tangent and the tangent of half its complement. 

Ex. 9. Prove that, in any plane triangle, the base is to the 
difference of the other two sides, as the sine of half the sum of 

12 


4 


4 . . hs oe eae , ae ™ 


é 


i830. ANALYTICAL PLANE TRIGONOMETRY. 
the angles at the base, to the sine of half their difference: also, 
that the base is to the sum of the other two sides as the cosine 
of half the sum of the angles at the base, to the cosine of half 
their difference. 

Ex. 10. How must three trees, A, B, C, be planted, so that 
the angle at A may double the angle at B, the angle at B dou- 
ble that at C; and so that a line “of 400 yards may just go 
round them? 

Ex. 11. In a certain triangle, the sines of the three angles 
are as the numbers 17, 15, and 8, and the perimeter is 160. 
What are the sides and angles ? 

ix. 12..The logarithms of two sides of a triangle are 
2.2407293 and 2.5378191, and the included angle, is 37° 20’. 
It is required to determine the other angles, without first find- 
ing any of the sides ? 

‘Ex. 13. The sides of a triangle are to each other as the 
fractions 1, 1,1: what are the angles? 

Ex. 14. Show that the secant of 60°, is double the tangent 
of 45°, and that the secant of 45° is a mean proportional be- 
tween the tangent of 45° and the secant of 60°. 

Ix. 15. Demonstrate that four times the rectangle of the 
sines of two arcs, is equal to the difference of the squares of 
the chords of the sum and difference of those arcs. 

Kix. 16. Convert formule ¢, Chap. III, into their equiva- 

lent logarithmic expressions ; and by means of them and for- 
mule , Chap. III, find the angles of a triangle whose sides 
are 5, 6, and 7. 
Ex. 17. Being on a horizontal plane, and wanting to ascer- 
‘tain the height of a tower, standing on the top of an inacces- 
sible hill, there were measured, the angle of elevation of the 
top of the hill 40°, and the top of the tower 51°: then measur. 
ing in a direct line 180 feet farther from the hill, the angle of 
elevation of the top of the tower was 338° 45’: required the 
height of the tower. Ans. 83.9983 feet. 

Ex. 18. From a station P there can be seen three objects, 
A, B, and C, whose distance from each other are known, viz. 
AB=s800, AG= 600, and BC=400 yards. There are also 
measured the horizontal angles APC=38° 45’, BPC—22° 30’. 
It is required, from these data, to determine the three dis- 
tances PA, PC, and PB. 

Ans. PA=710.193, PC=1042.522, PB=934.191 yards. 


rs *. 
mee 
| “. 
ee 


SPHERICAL TRIGONOMETRY. 


Having demonstrated in the treatise on Spherical Geome- 
try, several important properties of the circle of the sphere, 
and of spherical triangles, we shall now proceed to deduce 
various relations which exist between the several parts of a 
spherical triangle. These constitute what is called Spherical 
Trigonometry ; and enables us, when a certain number of the 
parts are given, to determine the rest. The first formula 
which we shall establish, serves as a key to the rest, and is to 
spherical trigonometry what the expression for the sine of the 
sum of two angles is to plane trigonometry. 


CHAPTER I. ' ra 


1. To express the cosine of an angle of a spherical triangle 
in terms of the sines and cosines of the sides. 


Let ABC be a spherical triangle, O 
the centre of the sphere. 

Let the angles of the triangles be de- 
noted by the large letters A, B, C, and 
the sides opposite to them by the corres- 
ponding small letters, a, 0, c. 

At the point A, draw AT a tangent 
to the arc AB, and Ad a tangent to the arc AC. 

Then the spherical angle A is equal to the angle TAt be- 
tween the tangents, (Spher. Geom. Prop. VIL.) 

Join OB, and produce it to meet AT in T. 

Join OC, and produce it to meet A? in ¢. 

Join T, t; 


132 SPHERICAL TRIGONOMETRY. 
Then, 


4h 
Gilet, AB=sec. c. 


OC 
Or = see. AC=sec. b 
aaatan. AB=tan. c 
At 
OG =tan. AC=tan. d 


Then in triangle TO¢ 


T? =OT’?+0? —20T . Ot cos. TOt 
| Ae ths ong OF OF cos. TO 
OC OG OG: OC * OC 
=sec.’ c+sec.” b— 2 sec. c sec, 6 cos. a, - (I) 
Again, in triangle T At 
Te =AT’?+Af — 2AT. At cos. TAt 
TeclAT Aci o¢ wah Ab msriy 9 
OG DCE" OG OE HGF A ae 
=tan.’ c-+tan.* b—2 tan.c tan. bcos. A. - - (2) 


Equating (1) and (2) 
tan.” c+tan.? b— 2 tan. c tan. b cos. A 
_==sec.’? c+sec."b — 2 sec. c sec. b cos. a 
—1+tan.*? c+1-+tan.? b—2 sec. c sec. b cosa 
.. —2 tan. c tan. b cos. A=2 —2 sec. c sec. b cos. a 


sin. c sin. b 1 1 
or, pay 2 Rare €O8. A =. iyo ——-., COs: & 
icos. c ” cos. b cos. c’ cos. b 
cos. a—cos. b cos.c 
cos. A— COS: &— ©08. 6 cos. c) 


sin. & sin. c 
Similarly we shall have, 


| 
cos. b— cos. a cos. c 
cos, B= ——_______ ———"" P(a) 
sin. @ sin. ¢ 
cos. e—ecos.a cos. b 
cos. C= —-——_—_ 


sin. @ sin. 0” 


2. To express the cosine of a side of a spherical triangle, in 
terms of the sines and cosines of the angles 
Let A, B, C, a, b, c, be the angles and sides of a spherical 
triangle; A’, B’, C’, a’, b’, c’, the corresponding qualities in the 
Polar triangle, 


Then by (a), 


SPHERICAL TRIGONOMETRY. 133 


sin. 0’ sin. c’. 


But (Spherical Geometry, Prop. X.), A’=(180°—a’), 
a'=(180° — A), b'=(180°— B), c’=(180°—C), 


cos.(180° — A) — cos.(180° — B)cos.(180°—C) 


C08. LED" a) ars Sain (180%) .sin(180°=0) 
Sere, GPE: A+cos. B cos. C) 
ee sin. B sin. C | 
Similarly, 
cos. b= Cisadua rake us a : ‘ (8-) 


sin. A sin. 


__ COS. C+cos. A cos. rea 
Shine ( sin. A sin. B J 


3. To express the sine of an angle of a spherical triangle, in 
terms of the sines of the sides of the triangle. 


By (a) we have, 
cos. a—cos. 0, cos. c 


sin. b sin. c 
Tittcenk: _cos. a—Ccos. b cos. c-+sin. b sin. c 
sin. 5 sin. c 
_©0S. a—(cos. b cos. c—sin. b sin. c) 
sin. 6 sin. ¢ 

cos. a—cos. (b+c) 

=" sin. 0 sin. c 

atb+c,. b+c—a 


cos. A = 


2 sin. Ji edo wie Chane, Eri. Ch, Ll.) 
sin. 6 sin. c 
b+-c 
Let s Peo eich 
22 
b+c—a 
. $—Gi= ari, 
oatiass. 
Se &: 
_a+tb—c 
2 sin. s sin. (s—a) 
woh ch COB: J Sees . 


iz? 


134 SHERICAL TRIGONOMETRY. 


Again, resuming the expression for cos. A, 
cos. 4 cos. c+sin. b sin. c—cos. @ 
1—cos. A = 
sin. 6 sin. ¢ 
__cos. (b—c)—Ccos. a 
~ sin. b sin. c 
atb—c . Str? 
LO ae 
sin. b sin. c 
_2 sin.(s—c)sin.(s—-b) 
sin. b sin. ¢ 


2 sin. 


(2.) 


Multiplying equations (1) and (2) 
1— cos.? Mecsed STOR IE Par Og Giri lie ie 
sin.” 6 sin.’ c. 


-. sin. AS———-2 — Vsin. s sin. (s—a) sin.(s—b)sin.(s—c) } 
sin.O sin.c 
Similarly, | 
2 


. 7 Es / Sin. § Sin. (s-——a) 5i0.($—o) Sit. (S—c 
sin. B= a aes See eae Po 


2 


__* __ “ysin. s sin (s—a)sin. (s-6)sin.(sc) 
sin,a@ sin.b 


sin. C= 


Now, by equation (1) we have, 
2 sin. s sin.(s—a) 


ICCB sin. b sin. c 
or 
A  2sin.s sin. (s—a) 
2 cos. 9 sin. bsin.c 
A Aes Mata) s sin. (s—a) 
aie OF sin. 6 sin. ¢ 
a 
wae. sin. $ sin. (s—) \ (72) 


2, sin. @ sin. ¢ 


C Ne s sin. (s—c) ~ 
cos. =e 


sin. a sin. 6 


Next, by equation (2), 
2 sin. (s—b)sin. (s—c) 


cos. A = - - 
sin. 0 sin: c 


SPHERICAL TRIGONOMETRY. 135 
Or, 
A ye 
2 fink nes 8) 
a sin. 6 sin. c} 
epee sin. (s—4)sin. sin. (s—6)sin. (s—c) ) 
aie 
pa ae Os) sin. b sin. ¢ 
Similarly, 
B sin. (s—a)sin.(s—c) 
ais Naor aimee f 
pia sin. (s—a)sin. (s—4) 
sin ANY aN sin. a sin. 6 J 


Finally, dividing a expressions (y.3 by those y. 2), we obtain, 


tan.“ ale! \ J a OS (s—b)sin. (s—c) } 


sin. $ sin.(s—da) 


_ Bo. /sin. (s—a)sin.(s—c) 
PEM dati. s ainoy 


tan =! / sin. (s—a)sin.(s—b) 


sin. ssin.(s—c) 


4. To express the sine of a side of a spherical triangle in 
terms of the sines and cosines of the angles. 


By (8) we have, 


‘gp. COS. A-+cos.B cos. C 
en sin. B sin. C 
cos. A-+-cos. B cos. C-+sin. B sin. C 


.. 1+cos. a= sin. B sin. C 
ane eet 
ee sib. sm..G 
pre Nh Tels <0 ao cer SB 
is oe) oie le 2. (PlaneTrig.Ch.IL.) 
sin. B sin. C 
Let 
Se gee vB celia 
and, 
pees A+B—C 


136 SPHERICAL TRIGONOMETRY. 
Hence, 
ee fe. 
eee 2 cos. (s’—C) cos. (s B) 


sin. B sin. C ster hi antes 


Resuming expression for cos. a, 
cos. B cos. C —sin. B sin. C-+cos. A 
sin. B sin C 
cos. (B+C)+cos. A 
Rs sin. B sin. C 
A+B+C B+C—A 
= 12 cos. ——5-—— cos. 


sin. B sin. C 
oe! cos. 8! cos. (8X) 


Pray Seep A sO) 
Multiplying Equations (1.) and (2.). . 


PLS Ass 4 cos. s’ cos. (s! -s) cos. ‘S —B) cos. (s’/——C} 
sin.” B sin.’ C 


1—cos.a@ = — 


2 
sin. B sin. C 


X /—cos. s’ cos. (s'/— A) cos. (s’/— B) cos. (s’/—C) ) 
Similarly, 


sin. a = 


sin. 6 = sin. A sin. C 518) 
x aN Ga ed at 7 ( 


X /__ cos. s’ cos. (s/— A) cos. (s’/— B) cos. (s‘/—C) 


By Equation (1) we have, 
2 cos. (s’/—— B) cos. (s’-—C) 


Wye nee sin. B. sin. C 
,@  2-os. (s'—B) cos. (s'— C) 
"92 COS. > sin. B sin. C 
ee cos. (s’_B) cos. (s'—C) ) 
Nid ae La V sin. B sin. C 
Similarly, P 
hed cos. (s/— A) cos. (s‘/—C) > (6, 2.) 
pee =f sin. A sin. C 
pile cos. (s‘/—— A) cos. (s‘/—— B) 
i Ne Say / sin. A sin. B 


SPHERICAL TRIGONOMETRY. 137 
By equation (2.) 


2 cos. s’ cos. (s’— A) 
Patna Page 1 or Sky 


l —'cos))-@ = : : 
sin. B- sin. C 
ake 2 cos. s’ cos. (s’/—A) 
ove = RY a ne 
bz 2 sin. B sin. C 
siniiee an A Fal0818' 208. (8 A), 
2 sin. B sin. C 
pe f= s’ cos. (s/— B) | 
1. = ae anaes Seen ae Oe. 
2 sin. A sin. C " oD 
sittiom = pape eS SPS EE) | 
2 sin. A sin. B ) 


Finally, dividing the expressions (6. 3.) by the expressions (6. 2.) 


tan oe — cos. s’ cos. (s’— A) ) 
2 cos. (s‘/—B)cos. (s’/—C) 
6 , / — cos. s’ cos. (s’/—B) | 
tan. > = ae al _— 
2 cos. (s/——A) cos. (s’/—C) i: 
c — cos. s’ cos. (s/— C) 
tg ee! A I re ge es 
igen cos. (s’/—A) cos. (s’—B) 


It is to be remarked that although the expressions (6. 1.) 
(5. 8.), (6. 4.), appear under an impossible form, they are in re- 
ality always possible. 

For by Prop. XVIII. of Spherical Geometry, the sum of the 
angles of a spherical triangle, is always greater than two 
right angles, and less than six right angles. 


», AF+B+C > 180° and < 540° 
A+B+C 
i —— or SSF and. 210% 
Hence, cosine s’ is always negative, and ... — cos. s’ is always 
positive. 


Again, if a’, b’, c', be the three sides of the polar triangle, 
since the sum of any two sides of a spherical triangle is 
greater than the third side: 


b+. c' > a! 
180°— B+-180°— C > 180° — A 
B+ C —A< 180° 


B+ - my tae 


| “a 
138 SPHERICAL TRIGONOMETRY. 


.. cos. (s’/— A) is always positive, and in like manner, cos. 
(s’ — B), cos. (s'-—-C), are always positive ; hence the above 
expressions are in every case possible. 

» 


5. The sines of the angles of a spherical triangle are to each 
other as sines of the two sides opposite to them. 

Taking the expressions (vy. 1.) and calling the common ra- 
dical quantity N for the sake of brevity: 


) 2N 

sin. & SS 
sin. 6 sin. c 

; 2N 

sin. B= 


sin. @ sin. ¢ 
Dividing the first of these by the second: 


sin, A sin. @ sin. c / sin. a ) 

sin. Bs sin. Osin.c ~~ ssin. D 
Similarly, 

sin. A __ sin. @ sin. b “ad sin. a («.) 

sin. C sin. c sin. 6 sin. Cc 


sin. B sin. b sin. a sin. b 
sin. C sin. csin.a@ sin. € J 


6. To express the tangent of the sum and difference of two 
angles of a spherical triangle, in terms of the sides opposite to 
these angles, and the third angle of the triangle. 


By (a) we have, 
cos. a — cos. b cos. c 


COS. A =. . ° - - - - - = - 1. 
sin. 0 sin. c ib 
And, 
cos. c— cos. a cos: 6 
cos. C = 
sin. a sin. b 
cos. c = cos. a cos.b+ sin. asin.bcos.C - - - (2) 


Substituting this value of cos. ein Equation (1.): 
cos. a— cos. a cos.’ b— cos. D sin. a sin. b cos. C 
cos. A =. 
sin. 6 sin. c 
cos. a (1 —cos.’ 6) —cos. 6 sin. a sin. b cos. C 
sin. 0 sin. ¢ 
cos, a sin. b— cos. b sin. a cos. C 


cats TRELINE ReMi ee ORR cine Eee iy ti (3.) 


sin. c 


2 i ‘ wail 
ye ‘ 
7. . . » 


pea Pk 
SPHERICAL TRIGONOMETRY. 139 


In like manner, substituting the value of cos. ¢ in Equation (2), 
in the expression for cos. B, we shall find, 

cos. 6 sin. a— cos. @ sin. 6 cos. C 

ea megane © + ee b00 oe) 


«€ 


cos. B = 


Adding equations (8) and (4) : 
cos. A + cos. B 


sin. a cos. b-+sin. 6 cos. a—(sin. a cos. b+sin. b cos. a) cos. C 


——s 
—— 


sin. c 
__ sin. (a-+b) — sin. (a+0) cos. C 
an sin. ¢ 
sin. (a+b) (1 — cos. C) 
re sinc  # BPR we >. * 
Again, by Equation (s) we have, 
sin. A + sin. a 
sin. B .. Speed 
. sin. A+sin. B __ sin. asin. 6 
sin. B a sin. b 


; 1 : sin. B 
*, sin. A=Esin. B* =< (sin. a = hed) — 
sin. b 


C 


sin. 
= (sin. asin. b) -— ae 
(sin. a+ sin. 6) cate (6.) 


Dividing Equation (6) by Equation (5), and taking first the 
positive sign: 


sin. A+sin. B sin. a+sin. 0 sin. C 
cos. A+cos. B ~ sin. (a+b) 1—cos. C 
erie Ste noe iar 2 sin. Ch Cos. Gi 
2 pe 2 C 
BE Append +o: a uc oe 
2 COS. 5) COs. heal 2 sin. 9 COS. 9 
a 
tan. 15 = ee Colm 
2 a+ 6b 2 
cos. — 


Again, dividing Equation (6) by Equation (5) and taking 


the negative sign. 


sin. A—sin. B sin. a— sin. D sin. C 


cos. At+cos.B sin. (a+4) 


1 — cos. G 


« 
- as 
ay 
f- . 
“4 


- 
ct 


- SPHERICAL TRIGONOMETRY. 


2 
2 sin. - 


sin. 


sin. 


a-— 


a—b 


2 
a+b 
COs. 


2 


COS. 


cot. 


COs: ===> 


2 


We have thus obtained the required expression, viz. 


Similarly, 


tan. 


tan. 


tan. 


tan. 


tan. 


A+B 


COS. 


COS. 


sin. 


sin. 


@—) 


COS. “ae. 


COs. 


sin. 


sin. 
COs. 


COS. 


sin. 


sin. 


cot. 


cot. 


cot. 


cot. 


cot. 


cot. 


vo | & 9 | D> vo] > 0} Q 


a | 


a 


/- ‘ oe. 


& 
SPHERICAL TRIGONOMETRY. 141 


7. To express the tangent of the sum and difference of twe 
sides of a spherical triangle, in terms of the angles opposite to 
them and the third side of the triangle. 


Let A, B, C, a,b, c, be the sides and angles of a spherical 
triangle, A’, BY, C’, a’, b’, c’, the corresponding parts of the po- 
lar triangle then by expression (2), ) 


g'—)p' 
ee oh aammere” 
cos vain j 
(2 
Therefore, 
yg 80° A) —(180°—B) 
tan, 180° +180°-5 2 “oot, (18O=9) 
2 ~_ (180°--A)-+(180°-B) “~" —_2 
cos. ———_____—_—— 
; 2 
z cos.( a 
a+b ee 8 c 
s0— = ¥ a - om 
tan. (180 > = ere cot. (90 5 
cos.( 180 eae 
; 2 
A—B 
. COS. Gc 
t at+b_ < fan. —— 
an Py bs 2 
cos. — 
a’ —— b! 
sin 7 C’ 
oa Ala Died “eot. — 
. 2 = a’+ b! 2 
d sin. oY ae 
Therefore, 
._ (180°—A)—(180°—B) 
(180°—a)-(180°-b) ee isha” + kan. a Le 8) 
Cee oT ~, (180°—A)+(180°-B) ei 
j sin. __ aa 
2 
. A—B 
Sil. = . 
a eee Bit 2 | tai PF 
2 _ A+B 2 
sin. —-—— 
4s 


142 SPHERICAL TRIGONOMETRY. 


We shall thus obtain another group of formule analogous 


to the last. ? 
(Aves ) 
Poe: cos res 
tan, = oe 
2 A+B 2 
COS. 
2 
As—B 
sin. —-—— 
a—b 2 c | 
Te ieee) Ai tan. ==> 
het o ye A Ae ba | 
sin. _ 
2 ) | 
cos pie | 
bre 20h gad 
ee ea OF ae 
COS. 5 | 
cage eee C. (¢') 
ey sin “IST a | 
tan. 5 = 7 tan. 9 
sin. 5 J 
sedate ) 
; oy b 
sie A+C 2 | 
COs. 5 
_A]Ee 
oes sin. ss 
eee ee tan. [=a 
tan. 5 HAS G 5} 
sin 5 J 


8. To express the cotangent of an angle of a spherical trian- 


gle, in terms of the side opposite one of the other sides and the 
angle contained between these two sides. 


By («) 
A cos. @a—cos. b cos. c 1 
COS. == e . = - - = e ea ae 
sin. b sin. c (1) 
and, 
cos. C— COS. a cos. b 
cos. C= ——————_ 


sin. @ sin. 0 


& 


_ 
a =s 
~ 


. SPHERICAL TRIGONOMETRY. | 143 


Hence, 
‘cos. c=cos. b+sin. a sin. db cos. C. 


Substituting this value of cos. ¢ in equation (1), it becomes 
cos. a—cos. a cos.” b—sin. a sin. b cos. b cos. C 


cos. A= — 
sin. 0 sin. c 
cos. a (1—— cos.” b)——sin. a sin. b cos. b cos. C 
r* sin. 6 sin. ¢ 
. cos. a (l—cos.* b)—— sin. a sin. b cos. 6 cos. C 
fs cos. A= i aS ee er ere e 
sin. b sin. c 
*. cos. A sin. c=cos. @ sin. b—sin. a cos. b cos. C 
But, 


‘ Sita 1A 
sin. c= — sin. a, by (¢ 
sin. A » by ), 


sin. C .4 ] ; 
cos. A rg sin. @=cos. a sin. b—sin. a cos. b cos C 


cot. A=cot. asin. d cosec. C—cos. d cot. C. 
Tn rlicn the cotangent of A is expressed 1 in the required 
manner. 


If in Equation (1), instead of substituting for cos. c, we had — 
substituted for cos. b, the value derived from the Equation. 
' cos. b— Cos. a COS. C 


ERs 


cos. B= 


sin. @ sin. c 
we should have found a value for cot. A in terms of a, c, B, or 
cot. A=cot. a sin. c cosec. B— cos. c cot. B 


Proceeding in like manner for the other angles, we shall ob- 
tain similaf results, and presenting them at one view, we have 


cot. A=cot. a sin. b cosec. C—cos. b cot. C ) ) 
=cot. a sin. c cosec. B — cos. c cot. B 

cot. B=cot. 8 sin. a@ cosec. C — cos. a cot. C 
=cot. b sin. c cosec. A — cos. c cot. A | (1) 

cot. C=cot. c sin. a cosec. B — cos. a cot. B 
=cot. c sin. b cosec. A —— cos. 6 cot. A 


J 


9. To express the cotangent of a side of a spherical tr tangle, 
in terms of the opposite angle, one of the other angles, and the 
side interjacent to those two angles. 


Let A, B, C, a, d, c; be the angles and sides ofa spherical triangle, 
and Al, BY, Cee es the corresponding parts in the polar 
triangle. 


144 SPHERICAL TRIGONOMETRY. 
Then by (x) 


cot. A’=cot. a’ ‘sin. b’ cosec. C’ — cos. b’ cot. C’ 
.°. cot. (180°—a)=cot.(180°—A) sin. (180°—B) cosec. (180°—c} 
— cos. (180°—B) cot. (180°— c) 
—cot. a= —cot. A sin. B cosec. e— cos. B cot. c 
*, cot. a=cot. A sin. B cosee. c+cos. B cot. c. 


Applying the same process to each of the expressions in (7), 
we shall obtain analogous results, and thus have a new set 
ei. 
- =cot. B sin. C cosec. a+cos. C cot. 


of formule: 
| (8) 
cot. c=cot, C sin. A cosec. b-+-cos. A cot. 
=cot. C sin. B cosec. a+cos. B cot. a J 


cot. a=cot. A sin. B cosee. c+cos. B cot. 
=cot. A sin. C cosec. 6+cos.. C cot. 


cot. b=cot. B sin. A cosec. c+cos. A cot. 


waa era 


By aid of the nine groups of formule marked, («), (9), (7); 
(6), (2), (2), (25, (n), (4), we shall be enabled to solve all the cases 
of spherical triangles, whether right-angled, or oblique-angled ; 
and we shall proceed in the next chapter to apply them. 


CHAPTER II. 


ON THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 


Spherical triangles, that have one right angle only, are the 
subject of the investigation of this chapter; those.that have 
two or three right angles are excluded. 

A spherical triangle consists of 6 parts, the 3 sides and 3 
angles, and any 3 of these being given, the rest may be found. 
In the present case, one of the angles is by supposition a right 
angle; if any other two parts be given, the other three may 
be determined. Now the combination of 5 quantities taken, 


oD. 4. ‘ 
3 and BG tg g 108 therefore ten different cases present 


themselves in the solution of right-angled triangles. 

The manner in which each case may be solved individual- 
ly, by applying the formulze already deduced, will be pointed 
out at the conclusion of this chapter; but we shall in the first 
place explain two rules, by aid of which the computist is en- 
abled to solve every case of right-angled triangles. These are 
known by the name of Napier’s Rules for Circular Parts ; 
and it has been well observed by the late Professor Wood- 


SPHERICAL TRIGONOMETRY. 145 


house, that, in the whole compass of mathematical science, 
there cannot be found rules which more completely attain that 
which is the proper object of all rules, namely, facility and 
brevity of computation. 


The rules and their descriptions are as follow: 
Description of the Circular parts. 


The right angle is thrown altogether out of consideration. 
The two sides, the complements of the two angles, and the 
complement of the hypothenuse, are called the circular parts. 
And one of these circular parts may be called a middle part 
(M), and then the two circular parts immediately adjacent to 
the right and left of M are called adjacent parts; the other 
two remaining circular parts, each separated from M the 
middle part by an adjacent part, are called opposite parts, or 
opposite extremes. 


This being premised, we now give 
Napier’s Rules. 


1. The product of sin. M and tabular radius=product of the 
tangents of the adjacent parts. % 

2. The product of sin. M and tabular radius=product of the 
cosines of the opposite parts. 

These rules will be clearly understood if we show the man- 
ner in which they are applied in various cases. 

Let A, B, C, be a sphericai triangle, right angle at C. 

Let a be assumed as the middle part. 

Then (90°—— B) and 0 are the adjacent parts. 

And (90°— c) and (90°—A) are the opposite parts. 

Then by rule (1) 

R Xsin. a=tan. (90° — B) tan. b 


s=cot.) B tana Do mists f= ping on a; uae = I(T) 
By Rule (2) 
R. sin. a=cos. (90°—— A) cos. (90° — c) 
S=Site A-SI. €C, et UU ap tea ess = (2) 


2. Let b be the middle part, 
Then (90°—A) and a are adjacent parts, 
Then (90°—c)and (90° — B) are opposite parts. 
Then by Role J, 
R. sin. b=tan. (90°—A) tan. a 
S=COLPAMtAt @ an = mg Swett we GB) 


12* 


146 _ SPHERICAL TRIGONOMETRY. 


And Rule II, 


and R. sin. b=cos. (90° — B) cos. se ggasi 


=sin. B sin. c 


3. Let (90° —c) be the middle part. 
Then (90° — A), and (90° — B) are adjacent parts, 


And b and a are opposite parts. 


Then, 


R sin. (90° — c)=tan. (90°——A) tan. taal, 
R. cos. c=cot. A cot. B 


And, 


R. sin. (90°— c)=cos. a cos. 8. 


R. cos. c=cos. a cos, 8 - 


4, Let (90° — A) be the middle part. 


Then (90° — c) and db are adjacent parts, 
And (90° —B) and a are opposite parts. 


Then Rule I. 

R. sin. (90°—— A)=tan. (90° — ce) tan. b. 
..., .R..cos, A=cot.‘c fen: 0" 

And Rule II. 

R. sin. (90°— A)=cos. (90°— 


B) cos. a, 


R. cos. A=sin. B cos. a 


5. Met (90°— 


B) be the middle part. 


8a od’) 


-. + @ 


Then (90°—c) and a are the adjacent parts, 


And (90°— 
Then Rule I. 


cos. B=tan. (90°—c) tan. a, 


=(ani-a-cote:c 


cos. B=cos. (90° A) cos. a 


=sin. A cos. b 


Collecting the above results, and making R= 


sin. a=cot: B tan. b* - 
sin. asin. A sine?-\ >. 
sin. b=cot. A tan.a- - 
sin. b=sin. Bsin.c - - 
cos.c=cot. Acot. BB - 
cos. c=cos. a cos.) - - 
cos. A=tan. 0 cot. c - 
cos. A=sin. B cos.a_ - 
cos. B=tan. acot.c - 
cos. B=sin. Acos. b- - 


1 


A) and 0 are the opposite parts. 


=. (9) 


= We) nsd(L0) 
, we shall have 
Sane ne (1) 
-- + Q) 
- => (8) 
- = = (4) 
== (5) 
-- + ©) 
re) 
-- = @) 
yo pai) 
- - = (10) 


ci 


4 SPHERICAL TRIGONOMETRY. 147 


| It now remains for us to show that these conclusions are 
accurate, and in accordance with the formule already deduced. 
Now by (a). 
cos. c— cos. a cos. 6 
sin. a sin. 
But when C=90°, then cos. C=0 
__ COS. C— COS. @ COS. b 
” sin. a sin. b 
. cos. c=cos. acos. b, which is formula (6) in the above table. 


cos. C= 


Again by (8) 
sin. @_ sin. 
: sin. c sin. C 
But when C=90° sin. C=1 


sin. a=sin. A sin. c, which is formula (2) above. 
Sihjerly, 


sin. 6 sin. B 
sin. c sin. C 
* .. sin. b= sin. B sin. c, which is formula (4). 


. Next since by (a) 
‘ cos. a—cos. 6 cos. c 


cos. A= : . ——; substitute for cos.c its value in (6). 
sin. 6 sin. ¢ - 
cos. a—cos. a cos. *b 

> sim 2 sin. c 
_cQs. a sin. 
"gin. € 

cos. a sin. b 

= sina 

sin. A 
.. sin.b=cot. A tan. a, which is formula (3.) 


——,; Be perinie for sin. c, its value as found in (2.) 


Again, a 
cos. a—cos. b cos. ¢ 
——— S: © substitute for cos. ds its value in (6.) 


cos. A= 


sin. 6 sin. c 
cos. ¢ 
—— — cos. bcos. ¢ 
» =cos. 0 
sin. 6 sin. c 
cos. c sin. 8 
~~ gin. c cos. b 
=tan. b cot. c, which is formula (7.) 


148 SPHERICAL TRIGONOMETRY. 
Again by (a.) cos. B 


__ cos. b—cos. a cos. ¢ 
«gin. @ SIN. € 

cos. )—cos. b cos.? a 
oF sin. @ sin. c 


substitute for cos. c its value from (6) 


cos. 6 sin. a N ; ’ r 
ee ee substitute for sin. c, its value from (4.) 
cos. 0 sin. a 
=~ sin 
sin. B 
Sin. a=cot. B tan. b, which is formula (1.) 
Again, cos. B * ail 
cos. b — cos. @ COS. € ‘ ; * 
ia aT a a ee substitute for cos. b, its value in (6.) ‘ 
* Se cos. aieOe. c | 
C08. a= J 
sin. @ sin. c 


COS. C SIN. & 
~ gin. C COS. @ 
=tan. a cot. ec, which is formula (9.) 


Next by (8.) 
cos. A+cos. B cos. C 


oe ae sin. B sin. C 
But C=90° .*..cos. C=0, and sin. C=1. y 
cos. A 


ane cos. O—— sin ar. 
sin. B 


.. cos. A=sin. B cos. a, which is formula (8.) 


Again, 
. B+cos. A cos. C 
COS. WG Sisbdeah So —.——— and when C=90°. 
sin. A sin. C 
_ COs. B ; 
~ sin. A : 
+, cos. B=sin. A cos. b, which is formula (10.) 
Lastly, 
COS. C-+cos. A cos. B din th; 
COR ere ne and in this case, 


SPHERICAL TRIGONOMETRY. 149 
__cos. A cos. B 
sin. A sin. B | 
=cot. A cot. B, which is formula (5.) 


We have thus proved the truth of the results derived from 
the application of Napier’s rules, and may therefore apply these 
rules without scruple to the solution of various cases of right- 
angled triangles. | 
Let us then take each combination of the two data, and de- 
termine in each case the other three quantities, adapting our 
_ formule to computation by tables. — 


1. Given A, B, required a, 8, ¢. 
, : cos. A 


R cos. A=sin. B cég,'a ... cos. dR = - + (i) 

= an ’ sin. F 
» _ »R cos. B=sin. A cos. b .. cos. b5=R — tae (2) 
. Racos. c=tot. A.cot R - = o-- - + = = (3) 


¥ 


2. Given a, b, required A, B, ¢, 

R sin. a =cot. B tan. d .. cot. B=R sin. a cot. b - (4) 

R sin. b =cot. A tan.a .. cot. A=R sin. b cot.a@ - (5) 

4 R cos. c=cos.acos.b6 = - = - = + =- = = (6) 


3. Given a, c, required A, B, d 
ie sin. @ 


R sin. a =sin. A sin. c .*. sin. A=R — - - (7) 
ba iIC 
COs.'C 
R cos. c =cos. a cos. b .*. cos. b5=R ~ @ - (8) 
Cos. a 
w © “Roos. B=tan.acot.c - - - apa toga tea (9) . 
4, Given 8, c, required A, B, a. 
: : : sin. 6 
R sin. d=sin. B sin. c .. sin. B=R- - (10) 
sin. ¢ 
- ~ COS. C 
™ Rcos.c =¢os.acos.b .. ads. a=R (11) 
' cos. b 
Rcos. A=tan.bcot.c - = - - = = = - = (12) 


5. Given A, c, required B, a, 8, 
Roos. A=tan. b cot. c .. tan.b=Reos. atan.c - (138) 
R cos. c =cot, A.cot. B.*. cot. B=Rtan. A cos. c* + (14) 
R sin. aysin. A sin. ca ee + 


6. Given B, c, required A, a, d. 
R cos. B=cot. c tan. a .*. tan. a=R cos. B tan. ¢ - (16) 


* i? 


150 SPHERICAL TRIGONOMETRY. 


R. cos. c =cot. A cot. B .:.: cot. A=R fan*B cos. & - (17) 
Resin. d=sin.Bsin.ec - - -~ = = ~ = = = (18) 


7. Given A, b, required B, c¢, a. 
R cos. A=cot.c tan. b.. cot. c=R cos. A cot. b - (19) 
R sin. b =cot. A tan.a.'. tan.a=R tan. A sin.b - (20) 
R cos. B=sin. Acos.b - - = - - - - - = (21) 


8. Given B, a, required A, c, b. 
R-cos. B=cot. c tan. @ .*...cot. c=R cos. B cot. a - (22) 
R sin. a =cot. B tan. b .. tan. 2=R tan. B sin. a - (23) 


BR cos. A=sin: B gos.a_ 0 © SF ae ae 24) 
9. Given A, a, required B, 8, c. 
ae es : cos. A & 
R cos. A=sin. Bcos. a .. sin. B=>R —— = - (25) 
COS. a 
R sin. a=sin. A sit.e «. sin c=R—— ~~ (26) 
sin. A : 
Rsin.b =cot.Atan.a@ - - - - - - - = ~ (27) 
10. Given B, b, required A, a, c. 
cos. B 
R cos. B=sin. A cos. b .. sin. A=R - (28) 
cos. b 
Risin. @—sin. Bisin: c .°: sini.c =F es - - = (29) 
sin. B 
R gin. a:=cot.Btan.G ~- ~. -~ -.-,- - .@= (30) 


CHAPTER IIL. 


ON THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 


The different cases which present themselves are contained 
in the following enumerations. 


1, When two sides and the included angle are given. 

2. When two angles and the side between them are given. 

3. When two sides and the angle opposite to one of them 
are given. 

4. When two angles and the side opposite to one of them 
are given. 

5. When three sides are given. 

6. When three angles are given. 


SPHERICAL TRIGONOMETRY. 154 


I. When two sides and the included angle are given. 


The remaining angles may be determi 
mula (¢.) a ate y be determined from the for- 


Thus, ig a, b, C, be given, A, B, c, required. 


a—b 
EB PME 20 C 
an. — a ak ° 9 
2 
. a—)d 
; A—B sin 9 ae C 
tan. = paar Se, 
i 
A x —B 
Whence 5 and a are known from the tables. 
Let ADB, 
: 2 
A+B. 
5) Q 
A=b+9 
B=t—o 


A and B being known, ¢ may be obtained from (:.) 
sin. to sinwf 
For ; == 
sin.@ sin. A 
sin. C 
sin. 
And, in like manner, if any two other sides and the included 
angle be given, the remaining parts may be determined. ? 


sin. C=SIN. @ 


II. When two angles and the side between them are given. 
The remaining sides may be determined from the formula(Z’.) 
Thus, let A, B, c, be given; a, 4, C, required. 
B 


7 
; eae COS. 2a i 
ati are ALB* 9.2 
COs. 
yh. 

icp s an Bs 

Px, sin. —> 
tan. — tan. —- 
2 _ _ A+B 2 

sin. joi ak 


Whence or" and = are known from the tables. 


152 : SPHERICAL TRIGONOMETRY. . % 
a+b 
Let aree’ 
R a—b Ps 
2 “= @ 
a=s' +o! 
b=s'—o! 


a and b being known, C may be obtained by (.) 
7 sin.C sin. c 
ue sin. A sin. @ 
sin. C=sin. A —< 
si 


n. a 
And, in like manner, if any two other angles and the in- 
cluded side are given, the remaining parts may be determined. 


III. When two sides and the angle opposite to one of them 
are given. 
The angle opposite to the other side may be found from 
formula (e.) 
Thus, let a, b, A be given, B, C, c, required. 
sin. B sin. d 
sin, A sin. a 
sinh eae abe 
sin. a 
The angle B being determined, the remaining angle C will 
be found from (¢.) 


a—b 


cos. —— 
A+B 2 C 
For tan. Te een cot. =r 


Hotels Sebacp. co tivrtolt 
cos.- 5 


The angle C being determined, the remaining side c will be 
found from (é.) 


For 


sin. c sin. C 
sin. a sin. A 
sin. C 


sin. c=sin. @ -— 
sin. A 


or c may be found from (2’.) 

And, in like manner, if any other two sides and the angle 
opposite to one of them be given, the remaining parts may be 
determined. 


— . 
4 © te 


a 


lal SPHERICAL TRIGONOMETRY. 153 


IV. When two angles and the side opposite to one of them 
are given. m 
The side opposite to the other angle may be found from for- 
mula (¢.) 
Thus, let A, B, a, be given; 8, c, C, required. 
sin. 6 sin. 
sin. a sin. A 
Sp eat sin. B 
i sin, b= sin. a — 
The side 6 being determined, the remaining side c will be 
found from (%') 


cos Sear 
a+b We ng 2 c 
: eat Ste agf et, ok aaa as 4; 1 ee 
Fox tan. 9 ALB 9 
COs. 
9 
A+B 
ty COS. aT ean 
tan 5) A-B an. 2 
COS. 5) 


The side c being determined, the remaining angle C will be 
found from (:.) 
Sai BIH. C 
For =a 
sin. A sin. @ 
sin. ¢ 


sin. C=sin. A — 
sin. @ 


or c may be found from (c.) 

And, in like manner, any other two sides being given and 
the angle opposite to one of them, the remaining parts may be 
determined. 


V. When three sides are given. 

The three angles may be immediately determined from any 
one of the formule (vy 1,) (7 2,) (7 38,) (vy 4.) 

The choice of the formula, which it will be advantageous to 
employ in practice, will depend upon the consideration already 
noticed in the solution of the analogous case in plane trigo- 
nometry. 


VI. When three angles are given. 
The three sides may be immediately determined from any 
of the groups of formule (6 1,) (62,) (63,) (0 4.) 


14 


, 


154 SPHERICAL TRIGONOMETRY. 


CHAPTER IV. 
ON THE USE OF SUBSIDIARY ANGLES. 


We have already explained in plane Trigonometry, the 
meaning of Subsidiary Angles, and the purpose for which they 
are introduced ; we shall now proceed to point out under what 
circumstances they may be employed with advantage, in 
Spherical Trigonometry. 

In the solution of case I., where two sides and the included 
angle were given, we first determined the two remaining an- 

les, and having found these, we were enabled to find the side 
also. It frequently happens, however, that the side alone is 
the object of our investigations, and it is therefore convenient 
to have a method of determining it, independently of the angle. 

Thus, for example, let b, c, A be given, and let it be required 
to determine a, independently of the angles B, c. 

By («,) we have 


cos. a—cos. b Gos. € 


es 


sin. b sin. c 
Whence cos. a=cos. A sin. d sin. c+cos. 6 cos. c. 
From which equation a is determined, but the expression is 
not in a form adapted to the logarithmic computation; we 
can, however, effect the necessary transformation by the intro- 
duction of a subsidiary angle. 
Add and subtract sin. 0 sin. c on the right hand side of the 
equation. 
Then cos. a 
=cos. A sin. b sin. c+cos. b cos. c+sin. b sin. c—sin. b sin. ¢ 
=cos. b cos. c-+sin. b sin. c+sin. b sin. c cos. A—sin. b sin. c 
=cos. () —c)— sin. bsin.c vers. A 
1— cos. a=1— cos. (b —c)-+sin. b sin. c vers. A 
vers, a=vers. (b—c)-+sin. } sin. c vers. A 


sin. 6 sin. c vers. A 
=vers. (b——c) ) 1+ jae Goo 


cos. A= 


(Redan 4—Sin- b sin. c vers. A 
vers. (b—c) 
*, vers. a=vers. (b—c){1+tan.? 6} 
=vers. (b—c) sec.’ 4 
from which a may be determined by the tables, 4 being known 
from the equation 
sin. b sin. c vers. A 


i 
rane vers. (b—c) 


SPHERICAL TRIGONOMETRY. 155 


In like manner in case II, where two angles and the in- 
cluded side were given, we first determined the remaining 
sides, and then we were enabled to find the remaining angle. 
Now, let us suppose that A, B, c, are given, and that we are 


required to find C independently of a and b. 
cos. C+-cos. A cos. B 


Brom: (3) ip: Cur ae sin. A sin. B 

Ae cos. C=cos. c sin. A sin. B—cos. A cos. B 

. 1—cos. C=1—-sin. A sin. B(1—vers. c)-++cos. A cos. B. 
=1-+cos. (A+B)-+sin. A sin. B vers. c. 


Feat A sin. B vers. c 


ied 
or °2 sin.’ 5=2 cos.” 


A+B sin. A sin. B vers. ¢ 
=2 cos.” ee ee 
2 2°cos 2 A+B 
“we z -. 


C A+B 
.. sin.2— = cos.’ sec.? 6 
Se 2 

If we assume : 
sin. A sin. B vers. c 


fan. 6= 


2 
2 cos 9 


In case IJ, where two sides and the angle opposite to one 
of them were given, we first determined the angle opposite to 
the other side, and then the remaining angles and the remain- 
ing side in succession. Now, let us suppose the side ¢, inde- 
pendently of the angle B and of each other, under a form 


adapted for logarithmic computation. 


To find C, we have (7.) 
cot. A=cot. a sin. b cosec. C — cos. b cot. C 


or cot.Asin.C=cot. a sin. b—cos. b cos. C 


or sin. C=cot. a sin. 6 tan. A —cos. 0 cos. C tan. A 
.. sin. C+cos. C cos. 6 tan. A=cot. a sin. b tan. A. 
sin. 4 
Let cos. 6 tan. A=tan. = 
cos. 4 
; sin. 6 : 
sin. C+ cos. C=cot. a sin. b tan. A 
cos. 4 
*, sin. C cos. 6+cos.C sin. 4=cot. a sin. 6 tan. A cos. 4 


. : sin. 4 
sin. (C-+é)=cot. a sin. b tan. ae SEP cae| 
f =cot. a tan. b sin. 4 
whence C is known, é being previously determined from equa- 
tion 


156 SPHERICAL TRIGONOMETRY. 


tan. 6=cos. b tan. A. 


To find c, we have from («.) 
cos. a—cos. b cos. c 


cos. A= : 
sin. 6 sin. c 
oe sin. c sin. b cos. A=cos. a— cos. b cos. ¢ 
; COs. a 
sin. c tan. b cos. A= — COs. Cc 
cos. b 
' COs. @ 
sin. c tan. b cos. Am——= — cos. c 
cos. } 
1 ‘e 6 
Let tan. b cos. A=tan. 6= “2 
_. cos. 4 
; sin. 4 cos. a 
sin. c—_—_—_ + COS. c= 
cos. 4 cos. b 
cos. a cos. 6 
cos. b 


whence ¢ may be found, é being previously determined from 
the equation : 
tan. 6=tan. b cos. A. 


In like manner, in case IV, when two angles and the side 
opposite to one of them were given, we first determined 
the side opposite to the other angle, then the remaining side 
and the remaining angle in succession. Now, let A, B, a, be 

~ given, and let it be required to determine c and C, independ- 
ently of 6 and of each other, and under a form adapted to lo- 
garithmic computations. If we take the formula (é.) 
cot. a=cot. A sin. B cosec. c+ cos. B cot. ¢ 
or cot.asin. c=cot. A sin. B+cos. B cos. c 


or sin. c=cot. A sin. B tan. a+cos. B cos. c tan. @ 
.. sin. c—cos.c cos. B tan. a=cot. A sin. B tan. a. 

, sin. 4 

Let cos. B tan. a=tan. 6— =" 
cos. 4 

‘ sin. 6 ; 
sin. c— cos. c=cot. A sin. B tan. @ 

COS. 

sin. (c —6)=cot. A sin. B tan. a cos. 4 


sin. 6 
cos. B tan. a 
=cot. A tan. B sin. 4 
whence c may be determined, 4 being previously known from 
equation tan. 6=cos. B tan. a. 


=cot. A sin. B tan. a 


SPHERICAL TRIGONOMETRY, Peer 


To find C, we have from (8) 


Soak ass 20% A-+cos. B cos. C 


sin. B sin. C. 
sin. B sin. C cos. a=cos. A+cos. B cos. C 
cos. A 


sin. C tan. B cos. a=- +cos. C 
cos. B 
sain Ctany Beeosa—cos C8S8 4 
cos. B 
in. 8 
Let tan. B cos. a=tan. 6=~" 
é cos. 4 
sin. Cx S12: pees cos. C=°0®: A 
Os cos. B 
cos. A cos. 4 
— cos, (C +46) =——— 
ae) cos. B 


whence C may be found, 4 being known from equation 
tan. é=tan. B cos. a. 
In the fifth and sixth cases, any one of the angles or sides 
required, may be found independently of the rest by the for- 
mule referred to. 


EXAMPLES IN SPHERICAL TRIGONOMETRY. 


Ex. 1. In the right-angled spherical triangle ABC, the hy- 
pothenuse AB is 65° 5’, and the angle A is 48° 12’; find the 
sides AC, CB, and the angle B. 

Ans. AC=55° 7’ 32" 
BC=42 32 19. 
angle B=64 46 14 

Ex. 2. In the oblique-angled spherical triangle ABC, given 
AB=76° 20’, BC=119° 17’, and angle B=52° 5’; to find AC 
and the angles A and C. 

Ans. AC=66° 5’ 36” 

angle A=131 10 42 

angle C= 56 58 58 

Ex. 3. In an oblique spherical triangle the three sides are 

G=Bioal 7?) pa=-114 foc on lees 
required the angles A, B, C. 

Ans. A= 62°39! 42” 

B=124 50 50 

C= 50 34 42. 

14* 


‘ 


APPLICATION OF ALGEBRA TO GEOMETRY. 


. 


ON THE GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTITIES. 


As lines, surfaces, and solids are quantities which admit of 
increase and decrease, like other quantities, they may, like 
others, be made the subjects of algebraical operations, either 
by their numerical representatives, or by symbols expressing 
such quantities. Itis only necessaryfor this purpose, that their 
representatives should possess values in relation to each other 
corresponding to the magnitudes of the quantities which they 
represent ; and that those values should be expressed accord- 
ing to the properties or relations of the lines, surfaces, or 
solids, to each other; subject to geometrical construction and 
algebraical notation. ‘J 

We are enabled, also, to express by lines, surfaces, and so- 
lids, the solutions furnished by Algebra. This is founded on 
the known properties of geometrical figures, corresponding to 
similar properties of the quantities algebraically expressed. 
In this view of the subject, all quantities under algebraic ex- 
pressions, may be conceived to be susceptible of some kind of 
geometrical construction. 

2. We will proceed to explain the manner of constructing 
those expressions, and representing, under a geometrical form, 
the conditions of an equation. This is called constructing the 
algebraic quantities. 

Ex. 1. Let it be proposed to construct such a quantity as 


b 
the following: — the value of the letters composing the quan- 


tity being known. 

From any point A draw two indefinite lines A 
AM, AN, making any angle with each other ; 
upon one of these lines AM take AB=c, and 
AD=a; then upon the line AN take AC=b. 
Having drawn the line BC, draw also DE par- 
allel to BC, this will determine AE as the va- 


lue of ee For the parallels DE, BC, give this 


proportion AB: AD:: AC : AB, (Prop. XIV. 
Cor. 2 B. IV. El. Geom.) orc: a::6: AE. M N 


B Cc 


. : % ald =" > 2 
_ APPLICATION, ETC. 159 
Me ab 
Therefore, AE=— Res MA te oe ey 


Hence, the line AE being the fourth proportional to the 
three lines represented by C, a, b, it may be used for the con- 


struction of the quantity—. 

2d. Hence, also if it were proposed to construct the quan- 
Ges ‘ G 
tity — it may evidently be done in a similar manner, since in 


ab 


this case. the lines and a would be equal, for if — = 
c 


then a=b. 
b 
3d. If it were proposed to construct eset it may be ob- 


served that the quantity may be resolved into the expression 
(a+d) b 
ct+d 


b 
shall have — to be constructed, which may be referred to the 


, hence representing a+d by m, and c+d by n, we 


former case. 
a’ 


—) 
4th. Let the quantity to be constructed ba — 3 it may 


be observed that a?— b’ is equivalent to (a+2) x @a); hence, 


2p? $e 
- fe WED, be represented under the form eo a) 


and we have only to find the fourth proportional to c, a+, 
a—b. 


bth. If the quantity to be constructed be —— oe , it may be put 


under the form Be ek md and having constructed ie in the 


manner just explained, we call the line given by this con- 


f ab Cc mc : 
struction, m; then a becomes —— which may also be 
constructed as above shown. 7 

6th. It will be presumed, therefore, that in order to con- 


ab . a’ b 
struct a - H 


3 


! a 
whence if we construct ‘ 


and represent its value by m, we 


mb 
may proceed to construct 7 


160 APPLICATION OF 


Thus the whole art consists in decomposing the quantity 
: ; : : a a’ 
into portions, each of which retains the form sereOL rene ts and 


although this process may appear sometimes difficult, yet we 
may easily arrive at the object proposed by employing trans- 
formations. e 


a’+b? 


7th. If, for example, is to be constructed, we may take 
°+5° a+am .. 
Le becomes an which may 
atam (at+m)a ai ‘ 
aa, Or Gan? * quantity easy to be con- 
structed after what has been said, when m and n are known. 
Now to determine m and 2, the equations l°=a’m, c’=an, 
: bs Cc 
give, m=—a and n= 


b'=a'm, and c’=an; then, 


be reduced to 


A which may be constructed by the 


methods already explained. | 

Thus, while the quantity is rational, that is, without radical 
expressions, if the dimensions of the numerator do not exceed 
those of the denominator except by unity, we may always re- 
duce the construction to the finding of a fourth proportional to 
three given lines. 

It sometimes happens, that quantities present themselves 
under a form, that renders recourse to transformations of no 
use ; this is when the quantity is not homogeneous, that is, when 
each of the terms of the numerator and denominator is not 
composed of the same number of factors ; when the quantity, 
for example, is such as ao 

But it should be observed, that we never arrive at a result 
of this kind, except when, in the course of an investigation, 
we suppose, with a view of simplifying the calculation, some 
one of the quantities equal to unity. If, for example, in 

3 2 3 
os we suppose b equal to 1, we shall have a But, 
as we never undertake to construct a quantity without know- 
ing the elements which we are to use for this construction, we 
always know in each case what is the quantity which is sup- 
posed equal to unity. We can always therefore restore it, 
and the above difficulty cannot occur ; because, as the number 
of dimensions must be the same in each term of the numera- 
tor, and also of the denominator, although the number of terms 
may be different in the one from what it is in the other, we 
restore in each term a power of the line, which is taken for 


ALGEBRA TO GEOMETRY. 161 


unity, sufficiently raised to complete the number of dimen- 

a LZ b +c 2 
a+ 

be the line which is taken for unity, we may write formula 

a+bd+ed 
Saab which may be constructed by making l’?=d m, 

c= dn, and a’ = d@’ p, which will change it into 

ip+bd*>+d'n 


sions ; thus, if we construct ; d being supposed to 


ad+dm 
dp+bd+dn (p+b+n)d 7 
ean ip Ll yiap or me gie. qyl ee quantity easily con- 
structed, when we have constructed the value of m, n, and p; 


b? 2 3 


namely, m = pt. Tn P= oe which is readily done after 


what has been said. 

Hitherto we have supposed that the number of factors, or 
the dimensions of each term of the numerator exceeds the 
number of factors, or the dimensions of the denominator only 
by unity. It may exceed that number by two or even three, 
but never by more than three, unless some line has been sup- 
posed equal to unity, or some of the factors do not represent 
numbers. 

3. When the dimensions of the numerator of the pro- 
posed quantity exceed by two the dimensions of the deno- 
minator, the quantity expressed is a surface, the construction 
of which can always be referred to that of a rhomboid. and 
consequently to that of a square. If, for example, the quan- 


: a’ 2 b 
tity to be constructed be pike 

a’+ab a+ab, 
it may be considered as a X relat Now po easily 


constructed, after what has been laid down, by considering it 


a+b f 
asa X Te Let us suppose therefore that m is the value 


* b 
of the line thus obtained ; then a X ie will become 


a xm. Now if we make a the altitude and m the base of a 
rhomboid, we shall have a@ x m for the surface of this rhom- 
boid, (Prop. VI, B. IV, El. Geom.) therefore, reciprocally, 


: ; : *b 
this surface will represent a X m, or ities eli 
ate 
: . @+be+d 
In like manner, the quantity Pern Ee may bereduced to 


atc 


162 APPLICATION OF 


a similar construction by making bc =a m, and d?= an; for 
a+amec+tand (et mar be) 
GC ais RS eae 

a+me+nd 
ae IR 2g refers itself to the preceding 
constructions, as also the values of m, n. Having found the 
value of this factor, if we represent it by p, we have only to 
construct a X p, that is, to make a rhomboid whose altitude is 
a and base p. 

4. Lastly, if the dimensions of the numerator exceed the 
dimensions of the denominator by three, the quantity expresses 
a solid, the construction of which may always be reduced to 
a parallelopiped. If, for example, we were to construct 
a’ b + a’ 6 ’ 

Taman ahha might consider this quantity as the same as 
a+ab a’ 


: tab, 
estates and, having constructed thiol in the man- 


ner already explained, if we represent by m, the line given by 
this construction, the question will be reduced to this, namely, 
to constructab xm. Nowab represents, as we have seen, 
a rhomboid ; if, therefore, we conceive a parallelopiped, having 
for its base this rhomboid, and for its altitude the line m, the 
solidity of this parallelopiped will represent a b X m, that is, 
@b+a@ B? 

atc 

5. What has been said will suffice for constructing any 
rational quantity ; we proceed now to rational quantities of the 
second degree. 


Ist. In order to construct Vad, let us draw ‘p 
an indefinite line AB, upon which we may take 
the part CA, equal to a, and the part BC, equal 
aC B 


it will then become 


Now the factor 


abx 


to b; upon the whole AB as a diameter, describe 
a semicircle, cutting in D, the perpendicular CD, 
raised upon AB at the point C ; then CD will be the value of 
Jab; that is, the value of Vg % is obtained by finding a mean 
proportional between the two quantities represented by a, 6. 
Indeed, we have 

AC :CD::CD:CB, 
or a~CD:: CD: 6; 
whence OD? =ab, or. CD = Jap 


Qnd. If we were to construct V3 ab + 8°, or which is the 


same thing, /(3 a + 6) 6, we should find a mean proportional 
between 3 a + 6 and 6. 


> 


ALGEBRA TO GEOMETRY. 163 


8d. In like manner, if the quantity to be constructed were 
va’ — 0’, 

we might consider this the same as V (a +6) (a—b); and then 

find a mean proportional between a + 6 and a—b, If the 

quantity were Va? + bc, and we make bc = am, then we shall 

have Va’ + am, or (a4 + m) a, which is constructed by find- 

ing a mean proportional between a + m and a after having 


i be 
constructed the value of m = > by the rules already given. 


4th. To construct Va? + 6’, we can in like manner make 
b? = am, and construct Va’ + am, in the manner just explained. 
But the property of a right angled triangle furnishes a more 
simple construction. If we draw the line AB, c 
equal to a, and at its extremity A erect a perpen- 
dicular AC, equal to b, joining BC, we shall have 
BC? = AB’ + AC? = a? + 6b’, and, consequently, 
BC = va’? + Bb’. 

5th. We can also, by means of a right angled 
triangle, construct Va? — 6? in a manner different 4 “ 
from that above given; draw a line, AB, equal c 
to a,and having described upon AB, as a diam- ON 
eter, the semicircle ACB, draw from the point A , 
a chord AC, equal to 6; then, if we draw BC, sig 
this line will be the value of Va?— 6’; for the triangle ABC 
being right angled, we shall have AB? = AC? + BC’; con- 
sequently, BC? = AB’ — AC’ = a?— 0’; therefore BC = 
Va? — 6°. 

6th. Hence, also, Va? + be admits of a different construc- 
tion from the above. Make be = m’, and construct Va? + m’?, 
as just shown, first finding for m a mean proportional between 
b and ¢, as indicated by the equation be = m’, which gives 
m= V be: 

7th. If there are more than two terms under the radical 
sign, the construction is to be reduced to one of the preceding 
methods by means of transformations. — If, for example, we 
have Va’ + bc + ef we may make be = am, ef =an, and we 
have Va? + am + an, or V(a + m + n) a, which may be 
constructed by finding 4 mean proportional between a and 
a+ m+n, after having constructed the values of m and n, 


be ef ; 
namely, m = —,n = —. We might, moreover, make 
a a 


be = m?, ef = n3, 


’ 


164 APPLICATION OF 


and then we should have to construct Va? +m? + n?. Now, 
when there are several positive squares contained under the 
radical sign, as Va? -+m*-++n?+p?+ &c. we may make 
Va?-+m?=h, | 

Vh2+n? =i, Vi2?+p? = k, and so on; and, as each of the 
quantities is determined by the preceding, the last will give the 
value of Va?+m?+n?+p?+ &e. In order to construct these 
quantities in the most simple manner, each hypothenuse is to 
be regarded successively as aside ; having, for example, taken 
AB = a, and raised the perpendicular AC E 
= e, we may join BC, which will be 1; 
then at the point C if we raise upon BC the 
perpendicular CD = ; and having drawn ¢ 
BD, which will be i, at the extremity D, 
we may raise upon BD the perpendicular 
DE = p, and BE will be 4, and equal to 
Va2+m?-+n?+p?. 

If some of the squares are negative, we 
may combine the method just given with A B 
that for constructing Va? — b2 

8th. Lastly, if the quantity to be constructed be of this form 


atts b+e 
Vd+e 
multiplying by Vd + e, will change it into a 


V (b+ c) (d+e); 
dyke 
then, by finding a mean proportional between 6 + ¢ and d+e, 


: : am . é : 
and calling it m, we have 7 pr which is easily constructed. 


The construction often becomes much more simple by set- 
ting out always from the same principles; but these simplifi- 
cations are derived from certain considerations which are pe- 
culiar to each question, and consequently can be made known 
only as the occasion presents itself. We will merely remark, 
in concluding, that although the construction of the radical 
quantities, which we have been considering, reduces itself to 
finding fourth proportionals, mean proportionals, and con- 
structing right-angled triangles, still we can arrive at con- 
structions more or less simple or elegant by the method em- 
ployed for finding these mean proportionals; we shall now, 
therefore, introduce two other methods of finding a mean pro- 
portional between two given lines. 

The first consists in describing upon the greater AB (see 
3d diagram to Art. 5) of two given lines a semicircle ACB, 


2% 


ALGEBRA TO GEOMETRY. 165 


and, having taken a part AD equal to the less, raising a_per- 
pendicular DC and drawing the chord AC, which will be a 
mean proportional between AB and AD; for, by drawing CB, 
the triangle ACB is right angled, (Prop. XIX, Cor. 2, B. III 
fil. Geom.) and consequently AC is a mean proportional be- 
tween the hypothenuse AB and the segment AD. (Prop. XVII, 
Cor. 5, B. LV, Hl. Geom.) f 


The second method consists in draw- D 
ing a line AB, equal to the greater given 
line, and having taken a part AC cqual 
to the less, describing upon the remain- 5, 


der BC.a semicircle CDB, to which if H 7 
we draw the tangent AD; this tangent is a mean proportional 
between AB and AC. (Prop. XX VII, B. IV, El. Geom.) 

It is evident, therefore, that rational quantities may always 
be constructed by means of straight lines, and radical quanti- 
ties of the second degree may be constructed by means of the 
circle and straight line united. 

As to radical quantities of higher degrees, their construc- 
tion depends upon the combination of different curved lines. 

We will now proceed to the consideration of questions, the 
solution of which depends either upon rational quantities or 
radical quantities of the second degree. 


Geometrical questions, and modes of forming equations there-~— 
from, and their solutions. 


6. The precepts usually given in algebra for putting questions 
into equations, are equally applicable to questions in geometry. 
Here, also, the thing sought is to be represented by some sym- _ 
bol ; and the equation is to be constructed in such manner, as 
to express the relations of the quantity represented by such 
symbol, in quantities that are known, or in those whose values 
are attainable ; and the reasoning is to be conducted by the 
aid of this symbol, and of those which represent the other 
quantities, algebraically, as if the whole were known, and we 
were proceeding to verify it; this method of proceeding is 
called analysis. 

Although in expressing geometrical questions by algebraic 
equations, we have more resources and more facilities accord- 
ing as we are acquainted with a greater number of the pro- 
perties of lines, surfaces, &c., still, as algebra itself furnishes 
the means of discovering these properties, the number of pro- 
positions really necessary is very limited. The two proposi- 
tions that similar triangles have their homologous sides propor- 
tional ; and, that, in a right angled triangle the square of the 


| 15 
“ 


* 
* 


PI 
166 APPLICATION OF 


ae 
hypothenuse is equivalent to the sum of the squares of the twe 
other sides, are the fundamental propositions and the basis of 
the application of algebra to geometry. 

But there are many ways of making use of these proposi- 
tions, according to the nature of the question, and there is al- 
ways a discretion to be exercised in the choice of the means, 
and manner of applying them; and this discertion can only 
be acquired by practice. 

When a geometrical question is to be resolved algebraically, 
it will be necessary to construct a figure that shall represent 
the several parts or conditions of the problem under considera- 
tion, and, if possible, get such expressions for the unknown 
quantities in terms of those that are known, as may easily be 
determined, according to the known properties of the figure. 
But if it so happens, that the required quantity can have no ex- 
pression which will render it available under the present con- 
struction, we may, frequently, by drawing lines having certain 
relations to the known parts of the figure and also to the un- 
known, so connect the known to those that are required, as to 
get available expressions for their values. Having proposed 
a figure as above, we may, by means of the proper geometri- 
cal theorems, proceed to make out as many independent equa- 
tions as there are unknown quantities ; and the resolutions of 
these will give the solution. 


PROBLEM I. 


The base BC, and the sum of the hypothenuse AB and perpen- 
dicular AC, of aright angled triangle being given, to determine 
the triangle. 


Let BC = 6, and AC = 2, and if AB-+ AC A 
be represented by s, then will the hypothenuse » 
AB be represented by s—a. 

Therefore, by the properties of the right angled 
triangle (Prop. XXIV, B. 1V, Hl. Geom.) we have 

ACG?_+ BC? = AB’ 
Or, Y+B = sv—2 sx + 2’, C B 
omitting 2? which is common to both sides of the equation, 
and transposing the other numbers we have, 
23x = sx — 

Or, x= s'—)? 


which is the value of the perpendicular AC; where s and d 


may be any numbers whatever, provided s be greater than b. 
(Prop. X, B. I, £1. Geom.) 


o'> 


a4 ' * gh i 


ALGEBRA TO GEOMETRY. 167 


If this quantity is to be constructed, it may be resolved 
(Art. 2, Hz. 4,) into the form 
__ (s+) (s—b) 
ve 2s 
and constructed as follows. From any point C, draw an in- 
definite line, CM, and perpendicular thereto another indefinite 
line, CN, 

Set off on CM, CB = 2s, and N 
take also CE = one of the factors 
of the numerator, as s+, and take 
CF = the other factor, viz: s—b, 
draw BEF and also EH parallel 
thereto, and CH will be the value 
of x required. 

In like manner, if the base and 
the difference of-the hypothenuse M 
and perpendicular be given, we 
shall have by putting d for the difference and the other letters 
b and x as before ; d+ for the hypothenuse. 

Whence we have, 
z+ Qde4+- 2 = 8 + 2 
/? «o= bd? 
2d 


B E C 


Which may be constructed as before 
PROBLEM II. 
To desribe a square in a given triangle. 


(By a given triangle is understood a triangle in which the 
construction is known, viz: one whose sides, angles, altitude, 
&c., are known.) 

It will be perceived that this question resolves itself into the 
determination of some point, G, in the altitude EF, through 
which a line AB, drawn parallel to HI, shall be equal to GE; 
we may, therefore, determine in algebraic expression, for AB, 
and also for GF, and put them equal to each other and we 


shall have a solution. 
E 
A \\. 
oral \ 


Be es D Fo 


I 


+ 


168 APPLICATION OF 


Let us therefore designate the known altitude EF by a, 
the known base HI by b, and the unknown line, GF by z; 
then will EG = a—z. 

Since AB is parallel to HI, we shall have 
BP: EG :: FP: GB:: El: BB:: HI: AB 

consequently, EF: EG:: HI: AB 


Or, a:a—x::b: AB, 
whence AB= Gesaes 
But, AB, GF i= .2, 
therefore, ap as e2 
and ab — be = ax 
Or, ab = ax + ber = (a + Db) g, 
h | Ug b 


In order to construct this quantity, it is necessary to find a 
fourth proportional to a+), b, and a (Art. 2,) which may be 
done as follows : 

From F to O apply a line FO-= a+4, that is, = EF + HI, 
and join EO; then, having taken FM = HI = 6, draw MG 
parallel to EKO, which, by its meeting with EF, gives the de- 
termination of GF, or the value of x; for the similar triangles 


EFO, GFM, give FO: FM:: FE: FG 
or, a+b6:b::a:FG 


ab 
therefore, FG = aig 


PROBLEM IIl. 


Given the base BC, and the angles B and C of the triangle 
ABC, to determine the altitude AD. 


(Angles are made to enter into an algebraic expression by 
the aid of lines employed in trigonometry, viz, sines, tangents, 
&e. Thus when it is said that an angle is given, it may be 
understood that the value of its sine or tangent is given.) 

_If we designate BC by a, and AB by y, we shall have CD: 
AD:: radius : tan. ACD, (Trigonometry) or if we designate 
the radius by r, au the tangent of the angle C by ¢, we have, 

syssrit 


ALGEBRA TO GEOMETRY. 169 


whence CD = ao . 
In like manner designating the tangent Co ae 
of the angle B by ¢’ we shall have, | nS ieee 
B . 


Drags r'ot 


whence BD = a 


Oe. 
But, CD+BD=BC=a 
therefore, a + 2 = ¢ 
whence by changing the construction we may present 
at t! | ; 
I FU + rt 


This expression is susceptible of greater simplicity, by in- 
troducing, instead of the tangents of the angles C, B, their co- 
tangents ; which, let us designate by q and q’; observing that 
the tangent is to radius, as radius to the cotangent, (‘Trigono- 
metry) and we shall have 

Liters cans and’ $7262 gee 
2 Emcee 7? 


% 
whence ¢ = —, and : 


substituting these values for ¢ and ¢’ in the former equations 
we have, 
ar. =. qq’ ar qq ar 


Schohum | 
From the above it may be perceived that when among 
quantities that are given those employed do not lead to results 
so simple as may be desired, it is not always necessary to 
commence the work anew in order to arrive at a more simple 
result; but it may be sufficient to express, by equations, the 
ratios of the quantities first employed to those which we 
would introduce, as we have expressed ¢ and ?@’ by the equa- 

2 


2 


r ; 
tions ¢ = — and?! = — by which a solution dependent upon 


q and q’ is obtained. 
PROBLEM IV. 


Given the three sides of a triangle ABC, to find the segments 
AD, DC, formed by the perpendicular BD, and also to find 
the perpendicular BD. 


15* 


i70 APPLICATION OF 


Let BD = y, CD = x, BC = a, AB= 8, AC = ¢, then AD 
or AG— CD = c—z. 
Hence, we have | , 
C+yY=@and?i—-2er+27+y= 0 
Let the second equation be sub- 
tracted from the first, and we have 
9 cre C = a’ — 2D’; 


whence, we have © | Ga Is 
a—P+c a—b’ 
—>= OD C 


Se = 
2c 2c 


—-+te A 


which may be resolved into 
2= fe b) fae 

By reference to article 2 it will be perceived that to obtain 
the value of 2, we have to find a fourth proportional to c, a+b 
and a— 6, to take one half of this and add it to 4e, or one half 
the side AC. 

Scholium. 

Several important conclusions may be drawn from this 
solution, some of which we will notice; showing at once, 
different modes of putting geometrical questions into equations, 
and how, by varying the propositions of these equations, new 
propositions may be discovered. | 

Ist. The equation 2 cz —c? = a? —}? is resolveable into 
ce (2a-——c) = (a + db) (a—D). 

Now since the product of the first two factors is equal to the 
product of the last two, we may consider the first two as the 
extremes, and the last two as the means of a proportion; hence 
we have 

c:a+b::a—b: 2xr—c, or e—(c—x); 

or,. AC: BC+ AB:: BC—AB:CD—AD 

2nd. If from the point C as a centre, and with a radius BC; 
we describe the arc BO, and draw the chord BO we have 

BD? + DO? = BO? ; 
now DO = CO — CD = BC— CD = a—z, 
therefore BO? = y? + a? —2axr + 2?; 
but we have found above y? + 22 =aq?; 
consequently BO? = 2 a? —2 ax = 2a (a—2). 


‘ a q? eS fe + c2 
Putting for x its value iggy wise 
since 2 ac — a? — ¢? = — (a? — 2ac +c?) =—(c —a)!, 


we shall have 


$2 ~— q?-—— p2 eines dy et 8 2 
BO: =2a(a+————) =2a(@e— ee +t 
ZC 2c 


Sauer” 


ALGEBRA TO GEOMETRY. «171 


= < (0 — (c — ay), 
Now, by considering c— a as a single quantity, we find 
bY — (c—a)! = (b+ c—a) (b—c +a), 
hence 
BO’ = = 6+ e—a) @—c +), 
which may be put under this form, 


BO = — (a+b +e—2a) (a+b +c—20). 


If, therefore, we designate the sum of the three sides by 2s, 
we shall have 


BO? = — (2s—2a) (2s—2ce)= 4— (s— a) (s—c). 


Letting fall from the point C upon OB the perpendicular Cl, 
we obtain from the right angled triangle CIO this proportion, 
CO: aes oR: sin. ed pat peonmne ey 


that is, 1 : + BO: -R:.sin. OCI; 
whence 141 BO = dows ol, or BO = iensiny OEE : 
R R 
4 a (sin. OCI)? 
consequently BO? = iat ese 


Putting these two valves of BO* equal to each other we have 
4a’ (sin. OCI 4a 
= 8) 0), 
or, dividing by 4 a, and fan the denominators to disappear, 
ac Gin. OCI)’ = R*(s—a) (s—e): 
that is, dividing by ac, putting R equal to 1, and extracting the 


square root, 
sin. OCL = / (s—@ S—s) 


ac 
which agrees with a formula, in Trig. 
3d. We may, from the equation y? + 2? = a?, deduce the 


following: y? = a? —x? = (4+ 2) (a—2), putting for x its 
value, as found in the problem we have 


a—b+ec b— qa?’ 
y= (ct Qe ) (a+ Qe ) 
= (Soe eS) eee aes) 
oe Qc oy Sey Rl SE 
(a + c) ver) é oa) 
={ 2c 2c 


172 APPLICATION OF 
r & tie 30) (ai+ saat) (” +c—a) (b—c+ 2) : 
i 2c 2c ; 
consequently, 


4 c2y? = (a+¢+4+ 6) (a+c—b) (6+c¢—a) (b6—c+a) 
=(a+6+c) (a+b +ce— 2b) (atb+ce— 2a (a+b+e — 2c) ; 
or, designating the sum of the three sides a + 6 + ¢ by 2s, 

4 c?y2=2 s (2s —2b) (2s —2a) (2s—2e) 
= 16 s (s—b) (s—c) (s—a), 
or, dividing by 16 and taking the square root, 


cy § qe Vy @ an.) et ee 
3 = Vs (s—8) (s—e) s—a) 
But ~, or ees is the surface of the triangle ABC. 


Hence, to find the surface of a triangle by means of the three 
sides, we must subtract each side successively from the half sum, 
multiply the half sum and the three remainders continually to- 
gether, and take the square root of this product; which agrees 
with Prop. XL, B. IV, El. Geom. 


4th. The equations 2 cx —c? = a? —b? may be resolved 
as follows, 
b? = a? +c? —2 cx 
but if the perpendicular fall without the tri- 8 


angle as in the present diagram, AD will © 

then be c + 2 instead of c — x, hence, desig- 

nating the sides as before, we have y? + x? 

= a’, and y? + c? + 2.cx+a?=0?, the first 5 


subtracted from the second gives c? + 2ca = - 


= 6? —a?, or c (c + 2x) = (b+ a) (b—a); 
whence, c:b+a::b—a:ct 2 
Now, c+2z,ore+c+a=CD+AD; 
consequently, AC: AB+ BC:: AB—BC:CD+ AD 
5th. The same equation c? + 2 cx = 6? —a?, may also be 
put under the following form 

b? = a? + c? + 2 cx, which answers to the last figure. 
comparing this with the equation, 

b? = a? + 2c —2 cx, which answers to the former fig- 
ure, we observe that b?the square of the side AB opposite to 
the acute angle C, is less than the sum of the squares of the 
other two sides a? +c? by 2 cx; on the contrary, the square 
of the side AB opposite the obtuse angle, (see last figure,) is 
equal to a? + c? + 2 ca, that is, greater than the sum of the 
squares of the other two sides, by 2cx, which agrees with 
propositions XX VI and XX VI, B. IV, El. Geom; by these 

propositions we may determine when the angles of a triangle 


ALGEBRA TO GEOMETRY. 173 


are to be calculated by means of the sides, whether the angle 
sought, be acute, or obtuse. 
6th. The two equations 62 = a? + c? —2 cx 
and 63 = a§ +07 +2 cx 
confirms the theory of positive and negative quantities, for it 
is plain that the segment CD takes different directions, accord- 
ing as the perpendicular BD falls within the triangle or with- 
out it. In these two equations the term 2 cz has, in fact, con- 
trary signs. Hence, whatever result we obtain with regard 
to one of these triangles, we obtain that which belongs to the 
analogous case of the other by merely changing its sign of 
that part which takes a different direction on the same line. 
Now, since in the above theorem, respecting the surface of 
a triangle the segment CD does not come into consideration ; 
therefore, the proposition is equally applicable to all kinds of 
plane triangles. 


PROBLEM. V. 


Having the lengths of the three perpendiculars, EF, EI, EH, 
drawn from a certain point EK, within an equilateral trian- 
gle ABC, to its three sides, to determine the sides. 


Draw the perpendicular AD, and having ri 
joined EA, EB, and EC, put EF=a EI=4, 
EH=c, and BD (which is 3BC)=z. 

Then, since AB, BC, or CA, are each=2z, 


we shall have, Prop. XXIV. B. lV. El. Geom, K 
AD= vy (AB’ — BD’)= vy (42° — 2’) = 32? 

rma af che 
And because the area of any plane triangle 8 a ial 


is equal to half the rectangle of its base and perpendicular, it 
follows that 
triangle ABC=1BC x AD=2xzy3=2' y3, 
BEC=iBCXEF=rxa. =aa, 
AEC=3ACxEI=«xb =r, 
- ALB=$ AB Mig X<c2.) en. 
But the last three triangles BEC, AEC, AEB, are together, 
equal to the whole triangle ABC, whence 
x’. /3=art+bet+cn. 
And, consequently, if each side of this equation be divided 
by 2, we shall have 
L/38=atb-+e, or 
a+b+c 
r= — 
V3 


174 APPLICATION OF 


Which is therefore, half the length of either of the three 
equal sides of the triangle. 


Cor. Since, from what is above shown, AD is=2z,/3, it fol- 
lows, that the sum of all the perpendiculars, drawn from any 
point in an equilateral triangle to each of the sides, is equal to 
the whole perpendicular of the triangle. 


PROBLEM VI. 


From a given point A, without a cirele BDC, to draw a straight 
line AK in such a manner that the part DH, intercepted in 
the circle, shall be equal to a given line. 


Since the circle BDEC a et 
is given, its diameter is 
supposed to be known ; f 
and, since the point A.is @-r-=---—7> 
given, we draw through 
the centre O the straight 
line AOC, the line AB is L 
to be considered as known, and consequently the line AC. In 
order to know how the line AE is to be drawn, we have only 
to determine what ought to be the magnitude of AD, that, 
when produced, the part DE should be equal to the given line. 
We will designate AD by z, AB by a, AC by 0, and the given 
line, to which DE is to be made equal, by c. 

Since the figure BDEC is a circle, the secants AC, AE, 
must be reciprocally proportional to the parts without the cir- 


cle ; that is, 

AC: AE:: AD: AB (Prop. XXXVI. B. IV. 
El. Geom.),or b:a+ce:: @© : a; 
whence 4 cxr = 2b; 
an equation of the second degree, which, being resolved, gives 

r= — 1 c£EV1 c+ab. 

of which the first value only, —ic+ Vici +ab, satisfies the 
question under consideration. 

In order to finish the solution, it is necessary to construct 
this quantity, which can be done without employing the trans- 
formations made known, art. 2. For this purpose, we draw 
from the point A the tangent AT, which, being a mean pro- 
portional between AB and AC, gives AT’=ab; the value of 
x therefore becomes | 

g=—1e+ViF+AT, | 
The radius TO being drawn, becomes a perpendicular at AT; 


ALGEBRA TO GEOMETRY. 175 


if then we take TI equal to 1c, by drawing AI, we shall have 
Al=Vv? c’+AT"; therefore, in order to obtain x, we have 
only to apply TI from I to R, and to describe from the point 
A. as a centre, and with the radius AR, the are RD, which 
will determine D, the peint sought; for Kutt 
AD, or AR=AI—IR=AI — TI=V1 7+AT?— ! cmz 

In order now to know what the second value of = signifies, 
namely, 3a 

g=—1c— v1 ¢?+ab, 

it must be observed that, as it is wholly negative, it can only 
fall in the direction opposite to that toward which AD tends. 
Let us see, then, if there be a question depending upon the 
same quantities and the same reasoning, which fulfils this con- 
dition. If now we suppose a and 0) negative, the equation 
x°+cxr=ab, undergoes no change; since, therefore, when the 
circle BDEC becomes B’D’E’C’, situated toward the left in 
the same manner that BDEC is toward the right, it follows 
that the solution of this case is contained in the same equation; 


the second value of z, or—}c~—VvV1ic’+ ab, belongs to the 
same case, and satisfies the same conditions; if, therefore, in 
the preceding construction, we apply IT from I to R’ on AI 
produced, and from the point A, as a centre, and with a ra- 
dius equal to AR’, we describe an arc cutting the circumfer- 
ence B’D’'E’C’ in E’, the point E’ will be such that the part 
intercepted, E’D’, will be equal to c. Indeed, 
AK/=AR’'=AI+IR’= V1 77+AT’+2 6, 

that is, AE’ is equal to the second value of z, the signs being 
changed. Now, since we apply this quantity in a direction 
opposite to that in which z extends, it follows that AE’ is in 
reality the second value of z. 

Hence, as the two circles are equal and situated in the same 
manner, the two solutions may both belong to the same circle, 
so that if we describe from the point A, as a centre, and with 
a radius AR’, the arc R’E, the line AE will also resolve the 
question ; indeed, it is evident that the point E, determined in 
this manner, is in the line AD, (obtained by the first construc- 
tion,) produced. But of the two solutions, furnished by alge- 
bra, the first falls on the right of the point A, and appertains 
to the point D of the convex circumference, while the second 
falls on the left, and appertains to the point E’ of the concave 
part of the circumference. 


a 


> 4 Fe, “ 


176 APPLICATION OF 


PROBLEM VII. 


Let it be required to find the direction of a given line AB from 
a point C, such, that its distance from the point A, shall be a 
mean proportional between its distance from the point B and 
the whole line. 


Let the given line AB be 
designated by a, and the dis- 
tance AC required by z; 
then BC will be a—zx; and, 


since the proportion required G ae 
is AB+s AC.:: AG.: CB, 
or EEG Me segaee sete eae 


we shall have 
m=O ON) OV LOLA 
an equation of the second degree, which, being resolved, gives 
r= —Latviadt+a’ 

In order to construct the first value of z, we must, accord- 
ing to what has been said, (Art. 5,) raise the point B the per- 
pendicular BD=} a; and, having drawn AD, we shall have 

AD=VBD?+AB?= V2 a’+a’; 
we have then only to subtract from this line the quantity 1 a, 
which is done by applying BD from D to O; then we shall 
have AO=V1 a?+a?—} a, that is, it will be equal toz. We 
then apply AO from A to C toward B, and C will be the point 
sought. 

As to the second value of z, namely, 


r= —1La—V1a@ +a’, 


¥ we apply BD from D to O' on AD produced, then we shall 
ave 


AO’'=1 a+ V1 a’+a? ; 
and, as the value of z is this quantity taken negatively, we ap- 
ply AO’ from A to C’ on AB produced in a direction opposite 
to that toward which z is supposed in the solution to extend ; 
and we shall have a second point C’, which will also be such, 
that its distance from the point A, will be a mean proportional 
between its distance from the point B and the whole line AB. 


Scholium. 1. We may observe that this question contains 
that of dividing a line in extreme and mean ratio; also the 


- construction which we have obtained, is the same as that 


given in the Elements of Geometry, (Prop. IV. B. IV.) But 
it will be perceived, that we are made acquainted with this 
construction by algebra, whereas in the Elements of Geometry 
we supposed the construction, and only demonstrated its truth. 


wg, 


“ 


» - 5 


ALGEBRA TO GEOMETRY. 177 


2. With a little attention to the course pursued in the pre- 
ceding questions, it will be evident that we have always taken 
for the unknown quantity a line, which being once known, 
serves, by observing the conditions of the question, to deter- 
mine all the others. This is the course to be pursued in all 
cases, but there is a choice with regard to the line to be used; 
there are often several, each of which has the property of de- 
termining all the others, if once known. Among these some 
would lead to more simple equations than others. The follow- 
ing rule is given to aid in such cases. 

3. If among the lines or quantities, which would, when taken 
each for the unknown quantity, serve to determine all the other 
quantities, there are two which would in the same way answer 
this purpose, and it would be foreseen that such would lead to the 
same equation, (the signs + and — excepted); then we ought to 
employ neither of these, but take for the unknown quantity one 
which depends equally on both ; that is, their half sum, or their 
half difference, or a mean proportional between them, or &c., 
and we shall always arrive at an equation more simple than 
by employing either the one or the other. 


4. The question we have resolved, (Prob. VI,) may be used 


to illustrate what is here said. In this question there is no 
reason for taking AD rather than AE, for the unknown quanti- 
ty; by taking AD for the unknown quantity z, we have z-+c 


for AE; and, by taking AE for the unknown quantity z, we - 


should have «—c for AD; and, as to the rest, the mode of 
proceeding is the same for each case; so that the equations 
differ only in the signs. If, therefore, instead of taking either 
for the unknown quantity, we take their half sum, and desig- 
nate it by 2, since their half difference DE=c is given, we 
shall have 
| AE=2+} c, and AD=z—} ca, 
whence, according to the proposition adopted in the first so- 
lution, 
(c+3 6) @—2¢)=ab 
or x — Fea, 
a more simple equation than the former, and which gives 
z=V1 c+ab; 
and, since AE=z+}-c, we have immediately 
, AE=1c+v! c?+ab, 
and AD=—ict yi @+ab, 
as before found. 


178 APPLICATION OF 


PROBLEM VIII. 


Let it be required to draw a right line BFE from one of the an- 
gles B of a given square BC, so that the part FE intercepted 
by DE and DC, shall be of a given length. 


Draw EG perpendicular A D 
to BE to meet BC produced 
in G, and from the angle E 
draw EH perpendicular to 
BG. 

Let BC or DC=a,FE=8, 
BF=y, and CG=z. 

Since the triangle EHG 
is similar to the triangle BCF 8 4 Mi 
and the side EH=the side BC, hence the hypothenuse EG= 
the hypothenuse BF. 

But BE’?+EG’=BG,, 
or 2y°+2bytVH=a+2ae+2° 
and because the triangle BCF and BEG are similar, 
BF: BC:: BG: BE 
or yah: aol: Fb 

hence, y+by=a+axr 
multiplying this equation by 2, we have 

2y?+2by=2a’+ 2ax. 

Subtracting the last from the former equation, we have 


E 
| 
| 


b= —a’?+a", 
or Y+a’=2z’, 
hence, r= V+? 


having the value of z, y may be found in the equation 
y?+by=a+ar 
completing the square y’+by+jb=a’+azr+ jb 


hence y= Va +ax+ib—hb 
Let DE=z, 
then ieee | We a 
“and yz=ab 
ab 
hence, | is 
y 


Scholium. This problem is susceptible of several modes of 
solution, but perhaps none more simple than the one here 
given, for most of the modes of which it is susceptible, involve 
powers and equations higher than quadratics. 


ig 


ALGEBRA TO GEOMETRY. 179 


EXAMPLES FOR PRACTICE. 


Ex. 1. To find the side of a square, inscribed in a given 
ici hose diameter is d. 1 
semicircle, w Fees lays 


Ex. 2. To find the side of an equilateral triangle inscribed 
in a circle whose diameter is d; and that of another circum- 
scribed about the same circle. Ans. id./3, and dV3 

Ex. 3. To find the sides of a rectangle, the perimeter of 
which shall be equal to that of a square, whose side is a, and 
its area half that of a square. Ans. a+ia/2 and a—tla/2 

Ex. 4. Having given the perimeter (12) of a rhombus, and 
the sum (8) of its two diagonals, to find the diagonals. 

Ans.1/2 + /54+/2 and 4—-v2 

Ex, 5. Required the area of a right angled triangle, whose 

hypothenuse is z** and the base and perpendicular x?” and 2’, 
Ans. 1.029085 

Ex. 6. Having given the two contiguous sides (a, 6) of a pa- 
rallelogram, and one of its diagonals (d), to find the other di- 
agonal. Ans. /(2a°+2b? — d’) 

Ex. 7. Given the base (194) of a plane triangle, the line 
that bisects the vertical angle (66), and the diameter (200) of 
the circumscribing circle, to find the other two sides. 

Ans. 81.86587 and 157, 48865 

Ex. 8. The lengths of two lines that bisect the acute angles 
of a right angled plane triangle, being 40 and 50 respectively, 
it is required to determine the three sides of the triangle. 

Ans. 35.80737, 47.40728, and 59.41143 

Ex. 9. Given the hypothenuse (10) of a right angled trian- 
gle, and the difference of two lines drawn from its extremities 
to the centre of the inscribed circle (2), to determine the base 
and perpendicular. Ans. 8.08004 and 5.87447 

Ex. 10. Having given the lengths (a, 6) of two chords, cut- 
ting each other at right angles, in a circle, and the distance (c) 
of their point of intersection from the centre, to determine the 


diameter of a circle. Ane 1(a?-L°) + 2c? 


Ex. 11. Two trees, standing on a horizontal plane, are 120 
feet asunder ; the height of the highest of which is 100 feet, 
and that of the shortest 80; where in the plane must a per- 
son place himself, so that his distance from the top of each 
tree, and the distance of the tops themselves, shall be all equal 
to each other ? 

Ans. 20./21 feet from the bottom of the shortest 
and 40,/3 feet from the bottom of the other 


180 APPLICATION OF 


Ex. 12, Having given the sides of a trapezium, inscribed in 

a circle, equal to 6, 4, 5, and 8, respectively, to determine the 
diameter of the circle. Ans. 35 / (130X138) or 6.574572 
Ex. 13. Supposing the town 4 to be 30 miles from 8, 8 25 

miles from c, and c 20 miles from a; where must a,house be 

erected that it shall be at an equal distance from each of them? 

Ans. 15.118578 miles from each, viz :- 

in the centre of a circle whose circumference passes through 
each of the three towns. | eu 


Ex. 14. In a plane triangle, having given the perpendicular — 


(p), and the radii (7 R) of its inscribed and circumscribing cir- 
cles, to determine the triangle. 

Qr / (2pr—4rR—7" 

Ans. The base rab ed me 


p—2r 


DETERMINATION. OF ALGEBRAICAL EXPRESSIONS FOR SUR- 
FACES AND SOLIDS. 


7. We have seen in the Hlements of Geometry, that surfaces 
depend upon the product of two dimensions, and solids upon 
the product of three dimensions; so that, if the several di- 
mensions of one or two solids, or two surfaces, which we 
would compare, have to the several dimensions of the other, 
each the same ratio, the two surfaces will be to each other as 
the squares, and the two solids as the cubes, of the homolo- 
gous dimensions ; and more generally still, if any two quanti- 
ties of the same nature are expressed each by the same num- 
ber of factors, and if the several factors of the one have to 
the several factors of the other, each the same ratio, the two 
quantities will be to each other as their homologous factors, 
raised to a power whose exponent is equal to the number of 
factors. If, for example, the two quantities were a 0 c d, a’ b’ 
c' d', and we had 

Gamt 0 = Ps aCe £ oe 
then we should have 


a'b a'c a’ d 
y= — cc = —, d' = —, 
a a a 
and consequently, 

a’*becd 
abed:dbeid::abcd: ban? 

a’ 4 

a : ae 

a’ ss gi 


What is here said is true not only of simple quantities; the 


——~ |=. . s 
7 g~s = “s 
- ’ J 


ALGEBRA TO GEOMETRY. 161 


same may be shown with respect to compound quantities. Let 
the quantities whose dimensions are proportional be 
ab +cd,a’bl' -c'.d'; 
since, by supposition, ) 
esa ost: ose B:dzd, 
we shall have 
t / 
ee ee 


/ 
a a 
and consequently 


mh 12 
ab+cd:iad bl te'd::ab+ed:~— + ves 


42 b 12 
1p acy ee 


::@(ab+cd):a"(ab+ cd), 
° a : a’, 


It follows, from what is here proved, that the surfaces of 
similar figures are as the squares of their homologous dimen- 


~ sions, and that the solidities of similar solids are as the cubes 


of their homologous dimensions ; for, whatever these figures 
and these solids may be, the former may always be considered 
as composed of similar triangles, having their altitudes and 
bases proportional, (Prop. XXIII. B. IV. Hl. Geom.,) and the 
latter as composed of similar pyramids, having their three di- 


~ mensions also proportional. (Prop. XXXII, Cor. 5, B. Il, £2. 


Sol. Geom.) 

It will hence be perceived, that quantities may be readily 
compared, when they are expressed algebraically ; and this 
may be done, whether the quantities be of the same or of a 
different species, as a cone and a sphere, a prism and a cylin- 
der, provided only that they are of the same nature, that is, 
both solids, or both surfaces. 


Let it be required to investigate the properties of a pyramid and 
and also of a frustum of a pyramid, 


Let h = the altitude, s = the greater base and s’ the smaller 
base, and h’ = the altitude of the vertical pyramid taken from 
the top of the frustum, 

then we shall have /s’: /s::h':h + h' or the altitude of 
the whole pyramid, and consequently, 

A+R) vi =h Vs=h vs! + hi vs' 
and hV/s—h' ~/se=hys 

dividing by Ys— vs’ 


16* 


» 


i82 APPLICATION OF 


we have h'= é le, 
J/Si— JSS 
whence h’ becomes known. 
Let now h + h’ be represented by k, and we may have 


SD hus 
from the first equation ie a 


then we shall have for the solidity of the whole pyramid-=*(1) 


1 hi 


the solidity of the small pyramid “ 


substituting for k its value, we have for the solidity of the 
hol 7 bal 
whole pyramid =~ 
asi, oigeti 
Bd sta ae 
h' 78/8 h' sfs—s' Js! 
NED J s! s') or 3 ( Js’ ) mpl SF thay ae 
putting for h’ its value found above we have, 
kif} y 8s J/s—s'V/s! 
: 3 (/s— Vs’) vs 
which being reduced gives, ; 
j 


hence the solidity of the frustum will be 


t 
3 (s + Vss' +s’) - - - += - - = (3) 

that is, the solidity of the frustum is equal to the sum of the 
greater base, the smaller base, and a mean proportional be- 
tween the two bases, multiplied by the altitude of the frustum ; 
which agrees with the proposition in geometry. 

And if the two bases are equal, viz: if s = s’ then the so- 
lid becomes a prism, and the expression will become 


Sis +s +s) or 4h (88) =hs - =i saga) 


that is, the solidity of the prism is equal to its base multiplied 
by its altitude. 

Let the lateral surface of the pyramid be used as an ele- 
ment in its investigation. 

To find the lateral surface of a frustum of a regular pyra- 
mid, having the two bases and slant height given, as well as 
the radius of the circle inscribed in the larger base. 

Let the larger base be called s, the smaller s’, the slant height 
h, and the radius of the circle inscribed in the larger base, r. 


The perimeter of the larger base will be = and the peri- 


eT 


ALGEBRA TO GEOMETRY. 183 


meter of the smaller base may be found from the following 
tee ne Neritacine patna, anailae 
VS: VS 3: tr ip vs e€ perimeter oO € smaller 


base : whence (= + : ) h = the lateral surface. 


aged ates 
Or we may investigate the surface of the frustum in con- 
nection with the whole pyramid of which it is a part. 


ssvs' hs : 
Thus ep es h: TF the slant height of the 


whole pyramid, which make = f, and the vertical pyramid cut 
from the frustum will be k —h. 


k 
Hence, we have 7's. Bes cases — the lateral surface of 
the whole pyramid. - - - - - - - - + = = - (5) 
And since the lateral surfaces of similar pyramids are pro- 
sk stk 


portional to their bases, we may make s: s’:: ah — the sur- 


face of the vertical pyramid cut from the frustum. 
Ske Site 

Hence aes te dee Coc os 
pans 9; 


or (s— s’) the difference of the two bases, multiplied by . the 


ratio of the slant height of the pyramid to the radius of the 
base, is equal to the lateral surface of the frustum. 


Mee 
It may be observed that the ratio ~ 1s constant whether ap- 


plied to the whole pyramid, to the pyramid cut off, or to the 
frustum ; and is such as would be represented by the sine of 
the angle formed by its slant side with the plane of the base. 

Cor. Whence we have for the lateral surface of the frustum 
of a pyramid, this rule: 

Multiply the difference of the bases by the sine of the angle 
which the slant side makes with the base, or by the ratio of the 
whole slant height of a perfect pyramid on the same base to the 
radius of the base, which will give the lateral surface. 

8. If x represent the ratio of the circumference of a circle 
to the diameter, a ratio whichis known with sufficient accu- 
racy for practical purposes (Prop. XIX. B. V. El. Geom.,) the 
circumference of any circle whose radius is 7, will be 277 - (1,) 
and its surface rr’ - - - - - - (2.) 
Hence it is evident that the areas of circles increase as the 
squares of their radii, r being always of the same value, the 
quantity *r’ depends on, and js proportional tor’ - (3.) 


8 


184 APPLICATION OF 


If h be the altitude of a cylinder the, radius of whose base 
is r, for its convex surface we shall have trh_ - - (4) 
for its solidity 77r*h - - - - - . 
and for the same reason we shall have mr”h’ for the 
solidity of another cylinder, whose altidude is h’, and 
the radius of whose base is 7’, hence the solidities of the two 
_eylinders are to each other as the altitudes multiplied by the 
squares of the radii of the bases. If their altitudes and radii 
of their bases are proportional, in which case the cylinders 
will be similar, we shall neve) 

hte hiieun Sa" 


he 
consequently =h’ < 


and the ratio PAT h 
rh:rh 
becomes oe 
or multiplying by x and dividing by 4, r* - - 6,) 
that is the solidities are as the cubes of the vail of their bases, 
as before shown in geometry. 
Also if the altitude of a cone is A, andr the radius of the 
base, its convex surface may be expressed by 


Cae gene AT 
[Vr + hig KG SPE OK BS ss ss (7) 


Or let k = the slant height of the cone, then will its lateral 
surface be 
rr Xk = rkr - - - (8 
which result will be also rb ee Tee we take © ar the area of 
the base, and increase it in the ratio ofr: k, viz. 
rhe 
or rkr 


rsksiaqr’: 


The solidity of the cone will be 
h_ «rh r 
cr X 3 3 - : - - - - (9) 
or if we multiply the convex surface of the cone by one-third 
of its distance from the centre of the base, (Prop. IX. B. IIL. 
El. S. Geum.) we shall obtain the same result. 
The distance from the centre of the base to the surface may 


h 
be expressed k:h:: r “the distance, - - (10.) 


hence the solidity will also be 


rh «rh 
rkrX raha as before. 


If h' = the altitude of a frustum of the cone, then may the 


h! 
slant height of the frustum be h : k : h’ ae the slant height 


ALGEBRA TO GEOMETRY. 185 


required, which call k’ ; let the radius of the pei base of 
re+2r! 
the frustum be 7’, then will the lateral surface he ax k! 
=rhiatrkig 0. Z ¥ - (Er 
2” = the greater base, and rr” may stelitoe base » (Prop. a 
B. Ill. Hi. S. Geom.) the solidity == (rr ar? + J gpp)-(12) 
2h! iad ( 

And since _ and ~ : 
cone on one or the other of the bases, and whose altitude is 
equal to that of the frustum, hence if one of those expressions 
is taken from that of the frustum, the remainder will express a 
conesected frustum. 


*h. 
Thus 1h (#r?-+ar?+ J apy") cae 1 aa = har? + Jerr?) - - (13) 


which i is a conesected frustum, eee a conical cavity on its 
larger base and 3 h (t1°+ Vay?) - - - > (14) 
expresses a eer da frustum the cavity of which is formed 
on the smaller base. 

Or if we multiply the convex surface by } its distance from 
the centre of either base, we shall have the solidity of a cone- 
sected frustum, whose cavity is taken from the opposite base 
(Prop. XI, Cor. B. II, El. Sol. Geom.) 

Thus rkr + r'k'a, formula (11) the expression of the lateral 


each expresses the solidity of a 


h 
surface, multiplied formula (10,) the distance of the surface 


from the centre of the larger base gives 
rhe + rvhkix (15) 


When the two bases of the frustum are equal, the coinsected 
frustum becomes a conesected cylinder, and r and 7’, and h’ 
becomes identical. Hence the uae es becomes 

= 29 ke | - - - - - (16) 
where & represents the altitude. 

9. Applying the same notation to express the sphere, we have 
for the surface of any sphere whose radius is r, 4rr° ; and 
4nr’X1r=—4ar'’, will be its Bona GErop: SL Bal Damiee 
Sol. Geom.) - - - - “pe apd) 

If the surface of a saonet zone is required, it may be ex- 
pressed by the product of the altitude of the zone multiplied 
by the circumference of the sphere; let h=the altitude, and 


we have 2arh for the spherical surface of the zone, - - (2.) 
The solidity of the sector of which this is the jodie base, 
is QerhXir=20crh, - - (3) 


Now, since ‘the sector CBAD Ges the diagt am to the Prob- 
lem on the 187 page,) may be considered to be made 


186 APPLICATION OF 


up of a cone CBD and a segment BDA; in order to express 
these portions, it will be necessary to find the area of the cir- 
cular section BD; for this purpose, since CP=CA — AP 
=r—h, and CB=r, we have in the right angled-triangle BPC, 
BP= V CB? — PC?= V7? — r’°+2rh—h?= V 2rh —h?, - (4.) 

The radius of the circular section BPD, hence the area of 
the circle will be 2rrh — ch’, - - n Yae  S GR 

And the solidity of the cone vill te 

(2erh — ah) Ae -h _ 2r *h ab h+ch - (6) 

Hence the solidity of the segment will be 

enth __Qer Be wh: 2, pa _ (0) 

Which may be resolved into rh?(r —h,)  - - (8.) 

Hence, the solidity of the segment is equal to the es otek of 
a circle, whose radius is the altitude of the segment multiplied 
by the radius of the sphere, minus a third of this altitude. 

10. Let 7’ be the radius of a sphere inscribed in a vertical 
polyedroid, and r the radius of the circumscribed sphere. 
Then (Prop. XVIII. B. III, Ei. Sol. Geom.) the surface of the 
polyedroid will be 2r’"X2r=4r'rr, -- - - - - - = (1) 

And since the surface of the circumscribed sphere is=4r’x, 
formula (1, Art. 9,) it follows that the surface of the polyedroid 
and that of its circumscribed sphere, are to each other as 7’: r, 
since those are the only variable quantities which enter into 
their expressions. 

If h be the height of any zone of the sphere, its surface, 
formula (2, Art. 9,) will be 27’7xh=2rhr, - - - - (2) 

The surface of the corresponding zone of the polved old. 
will be 27/7 x h=2r'ha, (P. XVII. Cor. B. Il. HLS. G) - (3.) 

And hence the surface of any zone or segments of the po- 
lyedroid and sphere made by the same perpendicular to their 
common axis, will be in the ratio of their whole surface, viz. 


7: 
TSP AMRE phils ty RS Ht TA ae oh HoeziqarR MA) 


The base of any segment of the polyedroid made by a plane 
passing through a circle common to the polyedroid and sphere, 
may be found from formula (5, Art. 9.) 

Let 2rrh — xh’, be the base of acommon segment of the 
sphere and polyedroid, and h the common altitude of the 
polyedroidal and spherical sector. 

The solidity of the ee ea tat sector will be 

Qrlhax— = = gr he a) oR! sts es (5.) 

And since the corresponding Ae sector is 17°hr, for- 

mula (3, Art. 9) it follows that the solidities of the corresponding 


ALGEBRA TO GEOMETRY. 187 


sector of these solids are as the ratio of: r’, as has been shown 
with regard to the convex surface. ‘The cone whose base is 
the base of the segment, and whose vertice is the centre of 
the polyedroid or sphere,formula (6, Art. 9,) may be expressed 
Q077*h — 38erh?+rh* 
— hence the segment will be 
“ar! "he —2arh+rri?—ch* 
| e=aah(ar er’ trh— hy ~~ - - = (&) 
If the polyedroidal segments consists only of a vertical cone, 
its solidity will be 2%rh — rh’) xX th=acrh’? —} rh’, subtract this 
_ from the spherical segment on the same base, formula (7, Art. 9,) 
arh?— nh’, and we have }rrh’?. - - - - - - = - (7%) 
~ which is the value of that portion of the segment of the sphere 
not included in that of the inscribed polyedroid, which is 
such a portion of the sphere as would be generated by the re- 


~ volutions of a circular segment as BD, about the axis B 
DC, passing through the centre of curvature, and per- 0f/ 


pendicular to the arc of the segment at the point of 
contact D 2 =! 


PROBLEM. 


It is required to find when the spherical segment and the cone 
composing @ spherical sector are equal to each other. 


Let ABED represent a sphere gene- 
rated by the revolution of a semicircle 
ABE about its diameter AE. The sec- - 
tor ABC, by this revolution, generates 
a spherical sector, which is composed 
of a spherical segment generated by the 
revolution of a semisegment ABP, and 
of a cone generated by the revolution 
of the right-angled triangle BPC. 

The solidity of the sector, formula (3, Art. 9) will be 247*h. 


The solidity of the cone, 277°h — arb, formula (6.) 


Now, in order that the cone may be eh to the segment, 
the sector, which is the sum of both, must be double the cone: 
hence, 27r°h=4arh=2rrh*?+2ch', 
dividing by 2, transposing, &c., 


188 APPLICATION OF 


rrh?—lerh-+idch’ 


dividing by xh rh=17r741h* 
transposing 1h'—rh=17r 
h?—3rh=—r* 


from which we obtain ds 
h=2re Vv Er? 
Of these two solutions it is evident that only h=2r— Vv 47% 


can satisfy the conditions of the question, since 3r+V4r* 
is more than 27, or more than the diameter of the sphere. 


EXAMPLES FOR EXERCISE. 


1. What is the solidity of the spherical segments of which 
the frigid zones are the convex surfaces, the altitude of each 
segment being 327 miles, and the radius of the base 15'75,28 
miles ? Ans. 1282921583 solid miles nearly. 

2. What is the solidity of the spherical segments of which 
the temperate zones are the convex surface, the radius of the 
superior base being 1575,28 miles, that of the inferior 83628,86 
miles, and the altitude 2053,7 miles ? 

Ans. 55021192817 solid miles nearly. 

3. What is the solidity of the spherical segment of which 
the torrid zone is the convex surface, the radii of the bases be- 
ing 3628,86 miles, and the altitude 3150,6 ? 

Ans. 146715018499 solid miles nearly. 

4. Having two vats or two tubs in the form of conical 
frusta, whose dimensions are as follows, viz.: the first has a 
base whose diameter is 3 feet, its altitude is 31 feet, and the slant 
height of its side is 4 feet: the diameter of the base of the se- 
cond is 31 feet, its altitude is 5 feet, and the curve surface is 
60 square feet, what must be the dimensions of one capable 
of containing as much as the other two, if*the diameter of the 
bottom and top, and the altitude are in the proportion of 
2 21 and 3. 

5. What is the difference in surface of a vertical hexedroid 
circumscribing a sphere whose diameter is 10, and the whole 
surface of a conesected frustum of a cone inscribed in the 
same sphere, and whose wanting base is 6, and perfect base 4? 


CONIC SECTIONS. 


There are three curves, whose properties are extensively 
applied in mathematical investigations, which, being the sec- 
tions of a cone made by a plane in different positions, as will 
be shown in another place, are called the Conic Sections. 
These are, 

1. The Parabola. 2. The Ellipse. 3. The Hypetboth. 


PARABOLA. 


DEFINITIONS. 


j. A Parabola is a plane curve, such, that if from any point 
in the curve two straight lines be drawn; one to a given fixed 
point, the other perpendicular to a straight line given in po- 
sition: these two straight lines will always be equal to one 
another. 


2. The given fixed point is called the focus of the parabola. 
3. The straight line given in position, is called the directrix 
of the parabola. 


Thus, let QAq be a parabola, S the fo- 
cus, Nv the directrix ; 

Take any number of points, P,, P., P;, 
cae in the curve; 

Join 8, P,;8,P,;8, P, 3... and draw 
PN SP NEL PS ING peteoperpendiéuiar’s of) “\ 
to the directrix; then ia 

SPUN! (Pls PNY, SPs 
PE SE 19, £ 

4. A straight line drawn perpendicular to the directrix. and 
cutting the curve, is called a diameter ; and the point in which 
it cuts the curve is called the vertex of the diumeler. 

5. The diameter which passes through the focus is called 
the axis, and the point in which it cuts the curve is called the 
principal vertex. 


17 


190 CONIC SECTION. 


Thus: draw N,P,W,, N,P,W,, 
N,P,W,, KASX, through the points n, 
P,, P,, P,, 8, perpendicular to the di- 
rectrix ; each of these lines is a dia- 
meter; P,,.P;, P,, A, are: the- ver-. K 
tices of these diameters; ASX is the 
axis of the parabola, A the principal 
vertex. N; 

6. A straight line which meets the 2 
curve in any point, but which, when produced both ways, 
does not cut it, is called a tangent to the curve at that point. 

7. A straight line drawn from any point in the curve, par- 
allel to the tangent at the vertex of any diameter, and termi- 
nated both ways by the curve, is called an ordinaie to that 
diameter. 

8. The ordinate which passes through the focus, is called 
the parameter of that diameter. 

9. The part of a diameter intercepted between its vertex and 
the point in which it is intersected by one of its own ordinates, 
is called the abscissa of the diameter. 

10. The part of a diameter intercepted between one of its 
own ordinates and its intersection with a tangent, at the extre- 
mity of the ordinate, is called the sub-tangent of the diameter. 

Thus: let TP¢ be the tangent at 
P, the vertex of the diameter PW. 

From any point Q in the curve, 
draw Qgq parallel to Té and cutting 
PW in v.. Through § draw RSr 
parallel to Té. 

Let QZ, a tangent at Q, cut WP, 
produced in Z. 

Then Qq is an ordinate to the di- 
ameter PW; Rr is the parameter of PW. 

Pv is the abscissa of PW, corresponding to the point Q. 

vZ is the sub-tangent of PW, corresponding to the point Q. 

11. A stright line drawn from any point in the curve, per- 
pendicular to the axis, and terminated both ways by the curve, 
is called an ordinate to the axis. 

12. The ordinate to the axis which passes through the focus 
is called the principal parameter, or latus rectum of the para- 
bola. 

13. The part of the axis intercepted between its vertex and 
the point in which it is intersected by one of its ordinates, is 
called the abscissa of the axis. 

14. The part of the axis intercepted between one of its own 


= » all 
we i 
ie ‘ 


~ PARABOLA. 191 


ordinates, and its intersection with a tangent at the extremity 
of the ordinate, is called the sub-tangent of the azis. 

Thus: from any point P in the 
curve draw Pp perpendicular to AX 
and cutting AX in M. Through $ 
draw LS/ perpendicular to AX. 

Let PT, a tangent at P, cut XA ' 
produced in T. 

Then, Pp is an ordinate to the axis ; 
L/ is the latus rectum of the curve. 

AM is the abscissa of the axis cor- 
responding to the point P. 

“ea the subtangent of the axis corresponding to the 

oint ©. 
: It will be proved in Prop. III, that the tangent at the princi- 
pal vertix is perpendicular to the axis; hence, the four last 
definitions are in reality included in the four which immedi- 
ately precede them. 


Cor. It is manifest from Def. 1, that the parts of the curve 
on each side of the axis are similar and equal, and that every 
ordinate Pp is bisected by the axis. 

15. If a tangent be drawn at any point, and a straight line 
be drawn from the point of contact perpendicular to it, and 
terminated by the curve, that straight line is called a normal. 

16. The part of the axis intercepted between the intersec- 
tions of the normal and the ordinate, is called the sub-normal. 

Thus: Let TP be a tangent at any 
point P. 

From P draw PG perpendicular to 
the tangent, and PM perpendicular to 
the axis. T 

Then PG is the normal correspond- 
ing to the point P; MG is the sub-nor- 
mal corresponding to the point P. 


192 CONIC SECTIONS. 
PROPOSITION I. THEOREM. 
The distance of the focus from any point in the curve, is equal 


to the sum of the abscissa of the axis corresponding to that 
point, and the distance from the focus to the vertex. 


That is, SP=AM-+AS. 
For 
SP=PN by Def. (1,) 
=KM *.* NM is a parallelogram. 
=AM+AK 


=AM+AS -.: AK=AS, by Def. (1.) 


PROPOSITION Il. THEOREM. 


The latus rectum is equal to four times the distance from the 
focus to the vertex. 


That is Li=4 AS. 


For, | j 
Li=2 LS, Def. (14.) cor 
=2 LN 
oe sath 


ane” AS—AR. 


PROPOSITION IH. PROBLEM. 
To draw a tangent to the parabola at any point. 


Let P be the given point. t 

Join S, P; draw PN perpendicular to ” 
the directrix. ) 

Bisect the angle SPN by the straight % 
line Tt. 

Tt is a tangent at the point P. 

For if Tt be not a tangent, let Tt cut 
the curve in some other point p. 

Join 8S, p; draw pn perpendicular to 
the directrix ; join S, N. 


CONIC SECTIONS. 193 


Since SP=PN, PO common to the triangles SPO, NPO, 
and angle SPO=angle NPO by construction, 

“. SO=NO, and angle SOP=angle NOP. 

Again, since SO=NO, Op common to the triangles SOp, 
NOp, and angle SOp=angle NOp. 

”, Sp=Np. 

But since p is a point in the curve, and pz is drawn perpen- 

dicular to the directrix, 
oe pN=pn. 

That is, the hypothenuse of a right-angled triangle equal to 
one of the sides, which is impossible, .*. p is not a point in the 
curve; and in the same manner it may be proved that no 
point in the straight line T¢ can be in the curve, except P. 

.. Tt is a tangent to the curve at P. 


«Cor. 1. A tangent at the vertex A, is a perpendicular to the 
axis. 


Cor. 2. SP=ST, 
For, since NW is parallel to TX 
“. angle STP= angle NPT 
= angle SPT by construction, 
SP=ST 


Cor. 3. Let Qq be an ordinate to the diameter PW, cutting 
SP in z. 
Then, Pz=Pv t Q 
For, since Qq is parallel to Té 
*, angle Prv=angle «PT 
= NPT by construc- 
tion, 
= Pvz interior oppo- T 
site angle, 


Pz=Pv 


Cor. 4. Draw the normal PG, (see diagram Prop. V.) 
Then, SP=SG, 
For since angle GPT is a right angle, 

angle GPT=PGT+PTG=PGT+S8SPT 
Take away the common angle SPT and there remains 
angle SPG=angle SGP 

ROSE E=SG. 


17* 


194 , PARABOLA. 
ily 


PROPOSITION IV. THEOREM. 
n ee 


The subtangent to the axis is equal to twice the abscissa. 
’ \ % " 


RP 


That is, —  MT=2 AM 


For, MT=MS+ST 
—MS+SP. Prop. IIL cor, 2... << 


=MS+SA+AM. Prop. I. A\F 2 
=2 AM. 


Cor. MT is bisected in A. 
PROPOSITION V. THEOREM. 


The subnormal is equal to one half of the latus rectum. That is, 


MG = Z 5 we denote the latus rectum by L. 


For, MG sc. —SM a 
SP—SM. Prop. III, cor. 4. 
AS + AM—SM. Prop. I. 
S+ AS + SM—SM 


AS 


el tl i Wl 


A 
2 
L 
2 Pr op. Il. 


PROPOSITION VI. THEOREM. 


If a straight line be drawn from the focus perpendicular to 
the tangent at any point, it will be a mean proportional between 
the distance from the focus to that point, and the distance from 
the focus to the vertex. 


That is, if SY be a perpendicular let fall from S upon Tt 
the tangent at any point P 
SP.; 


oh Ae fos boo. 
Join A, Y. Pt 
Since SP= ST, and SP is drawn per- yu 
pendicular to the line PT, 
le Yr. 
Also by per A ay A\S ™M 
A=AM 


-- Since AY an the sides of a tri- 
angle TPM proportionally ; AY is paral- 
le] to MP, 


~" > se > we ee ee 


PARABOLA. ‘ 198 


AY i is perpendicular to AM. * 

he the triangles SYA, SYT, are similar, 

BST : SY 3 RP SSA . 

or, SP:SY::SY:SA .. SP=ST by Prop. III, cor. 2. 
Cor. 1. Multiplying extremes and means, 

SY*=SP SA Ol rt 

Cor: 20,6 Pa SA tee 2 SY" 
Cor. 3. By Cor. 1, 


SA’ 
And since SA is constant for the same parabola, 
SP proportional to SY’. 
Cor. 4. By Cor. 1., 
AS .SP 


SY? 
- 48Y?=4AS.SP 
= L.SP. Prop. IL 


lI ll 


PROPOSITION VII. THEOREM. 


The square of any semi-ordinate to the axis is equal to the 
rectangle under the latus rectum and the abscissa. 


That is, if P be any point in the curve 


= L. AM. 
For, 
PM? = SP?-——-SM? (Prop. XXIV. B. IV. Ei. by 
Geom. 


= (AM=AS)*— (AM — AS)’ 
*SP=AM+AS(Prop. 1,) & SM=AM—AS , 
= 4 AS. AM. (Prop. X. and XI, B. IV. °G Mm 
El. Geum.) 
aR Prop. I. 
Cor. 1. Since L is constant for the same 
parabola PM? proportional to AM, 
That is, The abscisse are propotional to the 
squares of the ordinates. 


PROPOSITION VIII THEOREM. 


If Qq be an ordinate to the diam- 
eter PW and Pv, the corresponding 
abscissa, then, 


Qu = 45P x. Bo. 


Draw PM an ordinate to the axis. 'r 
Join S,Q; and through Q draw 
DQWN perpendicular to the axis. 


” ¥ 
. 


f 
196 CONIC SECTIONS. 


From § let fall SY pepibacienlat 
on the tangent at P. 
The triangles SPAY, QD», are similar. 
Qu: QD’: : SP? :S¥? 
Bcti se : SA, Prop, VI. Cor. 2. 
The triangles PTM, QD», are also similar ; 
a Ds SD : MT 
23 PM : PM .MT 
:: 4AS. AM:2PM.AM 
> 4AS : 2PM 
2PM. QD = 4AS. Dv 


But, 
— QN’= 4AS . AM—4AS. AN=4AS(AM~— AN) 
= 4AS.MN 
And, PM’— QN?= PM+QN) (PM—QN) 
= (PM+QN) . QD 


- (PM+QN) .. QD = 4AS. MN = 4AS. DP 
But, 2PM .QD = 4AS. Dv 
* (PM—QN). QD = 4AS. Po 
Or, QD? = 4AS. Pv 
4s Qv’? :4AS.Pv::SP:SA. 
Qv’?.= 48P . Pe: 


Cor. 1. In like manner it may be proved, that 
qv = 48P x Pv. 
Hence, Qu = qv; and since the same may be proved for any 
ordinate, it follows, that 
A diameter bisects all its own ordinates. 


Cor. 2. Let Rr be the parameter to 
the diameter PW. 
Then, by Prop. HI. Cor. 3. 
Pa abe. 
Ceol bys 
Now, by the Proposition, 
RVs 400 wry 


= 4SP’ 
., 4RV’ or Rr’? = 16SP? 
Rr= 4SP. 


Hence the proposition may be thus enunciated : 

The square of the semi-ordinate to any diameter is equal to 
the rectangle under the parameter and abscissa 

It will be seen, that Prop. VII. is a particular case of the 


present proposition. 


ELLIPSE. 


DEFINITIONS. 


1. Aw Evuirse is a plane curve, such that, if from any point 
in the curve two straight lines be drawn to two given fixed 
points, the sum of these straight lines will: always be the same. 

2. The two given fixed points are called the foci. 

Thus, let ABa be an ellipse, S and 
H the foci. Birks oft 

Take any number of points in the 
curve P,, PL, P,,---- 

Oi” Sie esl ees bay Dek es OA 
S.P,, H.P, ; - - - - then, a a 
SP, + HP, = SP, +.HP, = SP, 

+ HP, =---- 

3. If a straight line be drawn join- 
ing the foci and bisected, the point Ps 
of bisection is called the centre. 

4. The distance from the centre to either focus is called the 
eccentricity. . 

5. Any straight line drawn through the centre, and termi- 
nated both ways by the curve, is called a diameter. 

6. The points in which any diameter meets the curve are 
called the vertices of that diameter. 

7. The diameter which passes through the foci is called the 
axis major, and the points in which it meets the curve are 
called the principal vertices. 

8. The diameter at right angles to the axis major is called 
the axis minor. | 

Thus, let ABa be an ellipse, S_and 
Hi the foci. 

Join S,H; bisect the straight line 
SH in C, and produce it to meet at 
the curve in A and a. 

Through C draw any straight line 
Pp, terminated by the curve in the 

oints P, p. 

Through C draw Bod at right angles 
to Aa. 

Then, C is the centre, CS or CH the eccentricity. Pp is a 
diameter, P and p its vertices, Aa is the major axis, Bd is the 
minor axis. 


198 CONIC SECTIONS. 


9. A straight line which meets the curve in any point, but 
which, being produced both ways, does not cut it, is called a 
tangent to the curve at that point. . 

10. A. diameter drawn parallel to the tangent at the vertex 
of any diameter, is called the conjugate diameter to the latter, 
and the twodiameters are called a pair of conjugate diameters. 

11. Any straight line drawn parallel to the tangent at the 
vertex of any diameter and terminated both ways by the-curve, 
is called an ordinate to that diameter. 

12. The segments into which any diameter is divided by 
one of its own ordinates are called the abscisse@ of the dia- 
meter. 

13. The ordinate to any diameter, which passes through 
the focus, is called the parameter of that diameter. 

Thus, let Pp be any diameter, and 
Té a tangent at P. 

Draw the diameter Dd parallel to 
Tt. 

Take any point Q in the curve, 
draw Qq parallel to Tt, cutting Pp 
in v. 

Through S draw Rr parallel to 
Tt 


Then, Dd is the conjugate diameter to Pp. 

Qq is the ordinate to the diameter Pp, corresponding to the 
point Q. 

Pv, vp are the abscissee of the diameter Pp, corresponding 
to the point Q. 

Rr is the parameter of the diameter Pp. 

14. Any straight line drawn at right angles to the major 
axis, and terminated both ways by the curve, is called an 
ordinate to the axis. 

15. The segments into which the major axis is divided by 
one of its own ordinates are called: the abscisse to the azis. 

16. The ordinate to the axis which passes through either 
focus is called the latus rectum. 

(It will be proved in Prop. IV., that the tangents at the prin- 
cipal vertices are perpendicular to the major axis ; hence, de- 
finitions 14, 15,16, are in reality included in the three which 
immediately precede them.) 

17. Ifa tangent be drawn at the extremity of the latus rec- 
tum and produced to meet the major axis, and if a straight 
line be drawn through the point of intersection at right angles 
to the major axis, the tangent is called the focal tangent, and 
the straight line the directriz. 


, ELLIPSE. 199 


Thus, from P any point in the 
curve, draw PMp Risle pas ue 
to Aa, cutting Aa in M. 

Through S draw Li perpendicu- a 
lar to Aa. | 

Let LT, a tangent at L, cut Aa 
produced in T. 


Through T draw Nz perpendicular to Aa. 
Then, Pp is the ordinate to the axis, corresponding to the 
oint P. : 
! AM, Ma are the abscissx of the axis, corresponding to the 
point P. 

L1 is the latus rectum. 

LT is the focal tangent. 

Nv is the directrix. 

18. A straight line drawn at right angles to a tangent from 
the point of contact, and terminated by the major axis, is called 
a normal. 

The part of the major axis intercepted between the inter- 
sections of the normal and the ordinate, is called the subnor- 
mal. 


Let Tt be a tangent at any point r 
‘From P draw PG perpendicular N 
to Tt meeting Aain G. JIN t 
From P draw PM perpendicular , i 
to Aa. ; 4 


Then PG is the normal corres- 
ponding to the point P. ? 
MG is the subnormal correspond- 
ing to the point P. 


PROPOSITION 1. THEOREM. 


The sum of two straight lines drawn from the faci to any point 
in the curve is equal to the major axis. That is, if P be any 
point in the curve. 


SP + HP = Aa. 
For 


SP + HP = AS +AH Ls 
= AS + SH, 
And, > Def. 1. 
SP + HP = aS + aH : 
= 2aH + SH, > o 
.. 2(SP+HP) = 2 (AS + SH 
+ Ha) 


ie 


200 CONIC SECTIONS. 


Or, 
SP + HP = Aa. 
Cor. The centre bisects the axis 5 Tn for 
2AS+SH=2aH+SH. AS = aH 
- And, SC = CH by definition 3. AC»ra¥rau, 
Cor. 2. SP + HP =2 AC; "SP = 2 ACHP 
mP = 2 ACG—SP 
hence SP — HP = 2 AC .— 2 HF 


PROPOSITION Il. THEOREM. 
The centre bisects all diameters. 


Take any point P in the curve. 

Jon 8,P ;-H,P38,H4 

Complete the parallelogram SPHp 

Join C.p 3 CP; 

Then, since the opposite sides of 
a parallelogram are equal, 

SP = Hp, HP =Sp..SP+ PH 

i Sp pu 
. p isa point in the curve. 

Again, since the diagonals of paral- 
lelogram bisect each other, and since 
SH is bisected in C, 

*. Pp is a straight line, and a diameter, and is bisected in C. 

And in like manner, it may be proved that every other di- 
ameter is bisected in C. 


PROPOSITION Ill THEOREM. 


The distance of either focus from the extremily of the axis mi- 
nor is equal to the semi-axis major. 


That is, 
| SB or HB = AC. 
Since SC = HC, ard CB is com- 
mon to the two right-angle triangles 
SCB, HCB, R 


--SB = HB. 
But. 
SB + HB = 2 AC. Prop. I. 
- SB =\HB =. AC. A a 
Cor. 1. BC? = AS. Sa. 
For, 
6 


BC? = SB* — SC’ 


. s ie 
wl. é ELLIPSE: ©, 201 
AB? a SCs 
= (AC + SC). (AS—SC) 
= ANSdt Scat 


Cor. 2. The square ‘sor i eccentricity is equal to the differ- 
ence of the squares of the semi-axes ; 
For, SC? = SB — BC? 
= AC’— BC’. 


PROPOSITION IV. PROBLEM. 


To draw a tangent to the ellipse at any point. 


Let P be the given point. 
Join 8.P ; H,P produce SP. 
Bisect the exterior angle HPK 
by the straight line T¢. 
T?é is a tangent to the curve at P. 
For, if Tt be not a tangent, let 
Tt cut the curve in some other 
point p. 
Jon S,p; Hyp; make PR= PH, 
join p, K; H,1S cutting T¢ in Z. 
Since HP =PK, PZ common to the triangles HPZ, KPZ, 
and the angie HP Z=angle KPZ by construction, 
HIZ=K-Z, and the angle HZP =angle KZP. 
Pit since HZ=-KZ, Zp common to the triangles ees 
KZp, and angle HZp=angle KZp, 
pK sspit. 
But, since any two sides of a triangle are greater than the — 
third side, | 
Sp + pK > SK 
SSP PKR 
> SP + PH:.: PK=PH by construction: 
> Sp + pH, by definition 1, 
ve PAN or pide 
But we have just proved that pX=pH, which is absurd, 
.. p is not a point in the curve, and in the same manner it may 
be proved that no point in the straight line T¢ can be in the 
curve ees r. 
. Tt is a tangent to the curve at P. 


Cor. 1. Hence, tangents at A and a, are perpendicular to 
the major axis, and tangents at B and bd are perpendicular to 


the minor axis. 
18 


202 CONIC SECTIONS. 
Cor. 2. SP and HP make equal angles with every tangent. 


Cor. 3. Since HPK, the exterior angle of the triangle SPH, 
is bisected by the straight line Té, cutting the base SH pro- 


duced in T 
* ST2 HT 22 SP :-HP: 


t K 
P 


EL T 
re 


Oe a 


PROPOSITION V. THEOREM. 
Tangents drawn at the vertices of any diameter are parallel. 


Let Tt, Ww, be tangents at P,p, the vertices of the diame- 
ter'P’ Cp. 

JomuS Pas Hs 25.9% op, Hi: 
» Then, by Prop. Il, SH isa paral- 


a 
lelogram, and since the opposite | 7 
angles of parallelograms are equal, 

“. ang. SPH= angle SpH 
supplement of ang. SPH=supple- | * S 
ment of ang. SpH 8 
Rae 
P 


or, 
ang. SP T+ang. HPt=ang. SpW+ 
mck sag, SPT HP 
ut ang. = ang. t Rs - 
alae Spans pw | by Prop. IV. Cor. 2. 
Hence, these four angles are all equal, ss 
.. ang. SPT=ang. Hpw. 
And since SP is parallel to Hp, 
ang. SPp = ang. PpH, 
-. whole ang. TPp = whole ang. wpP, and they are alternate 
angles, 


—f 


Wem, 


.. Tt is parallel to Ww. 
Cor. Hence, if tangents be drawn at the vertices of any 
‘wo diameters, they will form a parallelogram circumscribing 


he ellipse. 


ee PY a ee are 7 


a 


ELLIPSE. ~ 203 
PROPOSITION VI. THEOREM. 


If straight lines be drawn from the foci to a vertex of any di 
ameter, the distance from the vertex to the insertion of the con- 
jugate diameter, with either focal distance, is equal to the semi 
axis, major. 


+ 


al 


That is, if Dd be a diameter conjugate to Pp, cutting SP 
in E, and HP ine, 
PT or Pes AC. 


Draw PF perpendicular to Dd, 
and HI parallel to Dd or Tt, cutting 
PF in O, 

Then, since the angles at 0 are 
rightangles, the ang. [PO=ang.HPO, 4 
and PO common to the two triangles 
pe sTO MANA ME 

Celt =etir: 

Also, since SC = HC, and CE is 

parallel HI, the base of a triangle SHI, 


ory kere Pak, 
Hence, 
2PKH=2EKI+2 IP 
; = SE + EI 4. IP + HP 
= SP + HP 
= 2AC 
a ACs 
Also, ang. PEe = ang. Pek. .”. PE = Pe, and 
Pe= AG, 


PROPOSITION VII. THEOREM. 


Perpendiculars, from the foci upon the tangent at any point 
intersect the tangent in the circumference of a circle, whose 
diameter is the major axis. 


From § let fall SY perpendicular on 
Té a tangent. 

Join 8, P: H, P; produce HP to meet 
SY produced in K. 

Join CY ; 

Then, since angle SPY =angle KPY 
(Prop. IV.) and the angles at Y are 
right angles, and PY common to the a-—< 
two triangles, SPY, KPY. 


° asd SP=PK 


i 
204 CONIC SECTIONS. 


And S¥¥k, x 
And, since SY=YK, and HC=CS, CY cuts the sides of 
the triangle HSK a aw ea 
™ . CY is parallel to HK. 
Also, since CY is parallel to HK, SY=YK, HC=C8, 
. CY=LHK 


Hence, a circle inscribed with Centre C and radius CA will 
pass through Y. 

And in like manner, if HZ be drawn perpendicular to T%, it 
may be proved that the same circle will pass through Z also. 


PROPOSITION VIII. THEOREM. 


Lhe rectangle, contained by the perpendiculars, from the foci 
upon the tangent at any point, is equal to the square of the 
semi-axis, minor. 


That is, Se 2 HZ=EC’. 
Let T¢ be a tangent at any point P. 
On Aa describe a circle cutting Tt € 
in Y and Z. 
Join S, Y; H, Z; 
Then, by the last Prop., SY, HZ are / 
perpendicular to Tv. i 
Produce YS to meet the circumfer- 
ence in y. 
Jon C, y; C,Z; a 
Since yYZ isa right angle, the seg- 9 ¥™ 
ment in which it lies is a semicircle, and 7. =. are stirs ex- 
tremities of a diameter. 
“. yCZ is a straight line and a diameter. 
- Hence the triangles CSy, HCZ, are in every respect equal. 
40 BY =HZ 
CA ee. i= AS. SA=BC’ Prop. III. Cor. 1. 


‘a fet ae 


~. 


- sisal 
ae ey Be 


PROPOSITION IX. THEOREM. 


205 


Perpendiculars let fall from the foci, upon the tangent at any 
point are to each other as the focal distance of the point of 
contact. 

aad 


That is, mSY: HZ::SP: HP. 
For the triangles SPY, HPZ, are TY 
nianifestly similar, 
SP SY.y HA SP AP. 


Cor. Hence, 


op end w WF ee i? 
SY aS Ve. A ap 
=BC?. as last Prop. 
=BC 550-80 . 
So also, oar, 
is Aga 9 aly Sp 
eer d wste Sas 


PROPOSITION X. THEOREM. 


i} a tangent be applied at any point, and from the same point 
an ordinate to the axis be drawn, the semi-azxis major is a 
mean proportional between the distance from the centre to the 
intersection of the ordinate with the axis, and the distance 
,com the centre to the intersection of the tangent with the axis. 


‘That is, CT:CA::CA:CM. 


a 


Since the exterior angle HPK is bisected by Tt, Prop. IV. 
ST : HT :: SP: HP. (B. IV. Prop. XVI. Hi. Geom.) 
ST+HT : ST—HT : : SP+HP: SP—HP 

18* 


¥ 


206 CONIC SECTIONS. 


or, 2CT :. SH. :: 2AC :SP—HP 
aT PEED ACK Mes OSH > SPtaps. . (1) 


But since PM is drawn from the vertex of the triangle SPH 
perpendicular on base SH, 
~, SM+tHM : SP—HP: : SP+HP : SM—HM 
or, © SH : SPHP, 2s: 2 ee 2 CM oe (2. 
Comparing this with the proportion marked (1,) we have, 
Spe 28! cM RE er my Ogre id tars Cae 4} 
or, CT eos, AG es Se Ms 


PROPOSITION XI. THEOREM. 


Lf a circle be described on the major axis of an ellipse, and if 
any ordinate to this axis be produced to meet the circle, tan- 
gents drawn to the ellipse and circle, at points in which they 
are intersected by the ordinate, will cut the major axis in the 
same point. 


Let AQa, be a circle described 
on Aa. 

Take any point P in the ellipse, 
draw PM perpendicular to Aa, 
and produce MP to meet the cir- 
cle in Q, join C, Q. 

Draw PT a tangent to theel- \ 
lipse at P cutting CA produced in T. 9 “~~~ 

Jon TQ. 

Then QT is a tangent to the circle at Q. 

For if TQ be not a tangent, draw QT’ a tangent to Q cut- 
ting CA in ‘T’. 

Then CQT” is a right angle. 

.. Since QM is drawn from the right angle CQT’ perpen- 
dicular on the hypothenuse. 

“, CT’: CQ:: CQ: CM. (Prop. XVII, Cor. 2.B. IV. E. G.) 

or ,CTHeCa 2: CAs CM as27CQ=CA; 

But, by the last proposition, 

CT’: €A ::CA SCM, 
ab Col =OTy, | 
which is absurd, therefore QT’ is not a tangent at Q; in the 
same manner it may be proved that no line but QT can bea 
tangent atQ, .. &e. 


oe as - > it i i i. 


: 


ELLIPSE. 207 


Cor. 1. Describe a circle on 
the minor axis. 
Draw Pm an ordinate to the mi- 
nor axis cutting the circle in q. 
Let a tangent at P cut the mi- * 
nor axis produced in @. 
Then, since Pm is parallel to AC, 
and PM to BC, 
Cia Gin? CT: aT 
=A CS eet 
::Cq’ :Cm? .-. the triangles CQT, Cmg are similar. 
Ps, BEE ao peed Ore CB::CB:CM. 
Which j is analogous to the property proved in the last pro- 
position for the major axis. 


Cor. 2. Join tg. 
We can prove as above, that ¢q is a tangent to the circle Bqb. 


PROPOSITION XIf. THEOREM. 


The square of any semt-ordinate to the axis, is to the rectangle 
under the absciss@, as the square of the semi-axis minor is to 
the square of the semt-axis major. 


That is, if P be any point in the curve, 
PM*SAM.: Mats 2 Berner 
Describe a circle on Aa, : 
and produce MP to meet it 
in Q. / 
At the point P and Q draw 
the tangent PT, QT, which« 
will intersect the axis in the \ 
same point T, (Prop. XI.) 
Let the tangent to the el- 
lipse intersect the circle inY,Z. 
Joie 3 HA; S¥ and HZ are perpendicular to Tt, 
(Prop. VIL.) 
Hence the triangles PMT, SYT, HZT, are similar to each 
other. 
oo RM Ss SB Y & t-PA PY 
and: PAs tar Zee: 2° WO sae 
. PM Se Zee MOP ave ALZ, 
or PM's (BO pe :: MIP eo TQta(Prop. XI.andProp. 


XVL B. IV. El. Geom.) 
>: QM’? : CQ’? -- MQT, MQC are si- 


milar triangles. 


We ee a ee 
oS *. ¢ ’ * 
s ) * i 


208 ' * CONIC. SECTIONS. . 


~-::AM.Ma: AC? 
*- PM?: AM. Ma : me BOT? AG: 


Cor. 1 
Let P,M ,P,M.,... be ordinates | 


to the axis from any points P,, P, 
Then by Prop. 


Bo Mo2 s AM, Mass. BC Sten. 
POM,’ : AM, . M,a:: BC? ws Oks 
(RM? .PM;? : AM, .M,a: AM, . Mga. 
That is, the square of the or ‘dinates to the axis are to each 
other as the rectangles of their abscisse. 


Cor. 2. By the fifth proportion in Frop. 
PM:QM::BC: AC. 


Cor. 3. By Prop. PM?: AM. Ma:: BC’: AC’. 
But as aa Ma=AC—CM, 
; (AC+CM) (AC — CM) : : BC’ AC? 
Pars AC? — CM? 


Cor. 4. Describe a circle on 
Bb, draw Pm, an ordinate to the 
minor axis cutting the circle in g. 


Then, Pm=CM , PM=Cm. 
Then by Cor. 3. 


Te ae 
ACG? Pmt: AC. 2 Cm : BC? 
Pm? le TA Gist: BCYCm : BC: 
crue +Cm) (BC —Cm) : BC? 
Bm . mb : BC? 
or, Pm > Bm. Hho AC? : BC? 


Which is analogous to the Storer proved in the proposition 
© for the major axis. 


Cor. 5. Pm: qm:: AC: BC. 


ae me ee ee 
° . *. » - * 
ELLIPSE. | | 209 


PROPOSITION XIII. THEOREM. 


The latus rectum 1s a third proportional to the axis major 


and minor. 
That is, Aa: Bb:: Bb: Ll . B. 
Since LS is a semi-ordinate to the 
axis, 
AC’: BC?:: AS .Sa@: LS’, Prop: XII, 
23 -BC*-«ES*, Prop: lly 4 a 
Cor. 1. 
AGE BEsesBO “SDS 
And, 
Aa:Bb::Bb =: Li. ce hal 


PROPOSITION XIV. THEOREM. 


The. area of all the parallelograms, circumscribing an ellipse, 
formed by drawing tangents at the extremities of two conju- 
gate diameters, is “constant, each being equal to the rectangle 
under the axes. 


Let Pp, Dd, be any two 
conjugate diameters, SROX 
a parallelogram circumscrib- 
ing the ellipse formed by 
drawing tangents at P, D, p, 
d; then Pp, Dd, divide ie par- 
allelograms SROX. into four 
equal parallelograms. 

Draw PM, dm, ordinates to the axis; PF perpendicular 
to Ad. 

Produce CA to meet PX i in T and Sd in t. 

Then, CT -GAe: 3 CA «GM 
And, Ct 3.QAneus GAs Gm 
fs CT: C¢ :: Cm: CM 
But, CT : Ct :: TM : Cm, by similar triangles. 
ve ATS .Cm?t. Cree 2a 
* OM.MT=Om s - - + 2. Q) 
ApainnOM.; CA: : CA: CT 
“ CM: CA:: MA: AT, dividendo. 
Or, CM: Ma:: MA: MT, componendo. 
_ AM. Ma=CM . MT—Cm? -.- =. (2) 
But, AC?! BC*: : AM. Ma (Cm’) : PM’. Prop. XII. 
+. SSAC BES: ee DMs 
Similarly, AC : BC :: CM: dm 
Or, “BC :.dm :: CA: CM 


210 CONIC SECTION. 


But, = COT 2. GAcsatwl Age CM 
“ag: Cote OU om 
Bat. PEs: sCT.. 4.dm.-«:Gd, 
for the triangle CdT=4 the parallelogram CPXd, 
Shy OO ot hace 4 OY Ue RY cL el Oe | 


*. rectangle PF . Cd=rectangle AC . BC 
or, parallelogram CX=rectangle AC . BC 
.. parallelogram SROX=4 AC. BC 
=Aa. Bd. 

Cor. By (2) Cm’?=AM . Ma 
=(CA+CM)’. (CA—CM) 
=CA’?—CM? 

“, CA?=CM?+Cm 
And similarly. CB’=PM?+dm’. 


PROPOSITION XV. THEOREM. 


The sum of the squares of any two conjugate diameters, is equal 
to the same constant quantity, namely, the sum of the squares 
of the two azis. 


That is, 
If Pp, Dd, be any two conjugate > B 
diameters, 
Pp? + Dd? = Aa + Bb’. 
Draw PM, Dm, ordinates the 


axis. 


Then, by Cor. to Prop. XIV, < Mf 
AC? + BC? = CM? + Cm? + 7 
PM? + Dm . ”, 


= CP? + CD: : 
*, 4AC?-+4BC?=4CP?-++4CD? 
Or, Aa?-+Bb?=Pp? +Da?. 


PROPOSITION XVI. THEOREM. 


The rectangle under the focal distances of any point is equal to 
the square of the semi-conjugate. 


That is. if CD be conjugate to 


Sb. eae Ot 
Draw SY. HZ, perpendiculars to 
the tangent at P, PF perpendicu- 
lar on CD. 
Then by similar triangles SPY, 
Pie 


ELLIPSE. 211 


Pr Sek seer base-SEa 
Or, SP : S¥ 22 AC : PF’. PE=AG) by Prap, VI 
ue aa epost <P, 
.SP. HP: SY.HZ:: AC’: PE’, 
sO D* 3 BC, Prop. XIV. 
But SY oHZ = BCR by Ev VILL. 
. SP. HP = CD*. 


PROPOSITION XVII. THEOREM. 


If two tangents be drawn, one at the principal vertex, the other 
at the vertex of any other diameter, each meeting the other 
diameter produced, the two tangential triangles thus formed, 

_ willbe equal. 

That is, 

trian. CPT = trian. CAK. 
Draw the ordinate PM, then, 
CM: CA: :CP:CK, by simi- 
lar triangles. 
But, COM : CAs CA LCL, Props Xs 
WEN SUP OOP TCR: 

The triangles CPT, CAK, have 
thus the angle C common, and the 
sides about the angle reciprocally 
proportional ; these triangles are 
therefore equal. 


Cor. 1. From each of the equal triangles CPT, CAK, take 
the common space CAOP; there remains, 


triangle OAT = triangle OKP. 


Cor. 2. Also from the equal triangles CPT, CAK, take the 
common triangle CPM; there remains, 
triangle MPT = trapez. AKPM. 


PROPOSITION XVIII. THEOREM. 


The same being supposed, as in last proposition, then any 
straight lines, YG, QE, drawn parallel to the two tangents 
shall cut off equal spaces. 


That is, 
triangle GQE = trapez. AK XG 
triangle rqik = trapez. AKRr 
Draw the ordinate PM. 
The three similar triangles 
CAK, CMP, CGX, 
are to each other as CA?, CM’, CG’, 


212 CONIC SECTIONS. 


trap. AKPM : trap. : AKXG::CA* —CM? : 
CA?—CG?, dividendo, 


But, PM’: QG :: CA?—CM?’: CA?—CG’, 
.. trap. AKPM : trap. AKXG:: PM? se 
But trian. MPT: trian. GQE:: PM? 5 ONG: 


.. the triangles are similar. 
.. trap. AKPM: trian. MPT :: trap. AKXG : trian. GQE. 
But, by Prop. XVII. Cor. 2, 

trap. AKPM = triangle MPT 
. trap. AK XG = triangle GQE 

And similarly, trap. AKRr = triangle rqi 
Cor. 1. the three spaces AKXG, TPXG, GQE, are all 

equal. 

Cor. 2. From the equals AKXG, EQG, take the equals 

AKRr, Egqr ; there remains, 

RrXG = rqQG. 
Cor. 3. From the equals RrXG, rqQG, take the common 
space rqguXG there remains, 
triangle vpQX = triangle vgR. 
Cor. 4. From the equals EQG, TPXG, take the common 
space EvXG ; there remains, 
TPvE = triangle vQX. 
Cor. 5. If we take the particular case in which QG coin- 
cides with the minor axis, 
The triangle EQG becomes the tri- 
angle IBC. 
The figure AK XG becomes the trian- 
gle AKC, 
.. triangle IBC = triangle AKC 
= triangle CPT. 


PROPOSITION XIX. THEOREM. 
Any diameter bisects all its own ordinates. 


That is, 
If Qq be any ordinate to a diameter CP, 
Qv = vq 

Draw QX, 92, at right angles to the major 

AXIS 5 

Then triangle vQX = triangle vgzx; Prop. 
XVIIL, Cor. 3. 

But these triangles are also equiangular ; 

*»| Qtr ng, 

Cor. Hence, any diameter divides the ellipse into two equal 

parts. ) 


ELLIPSE. 213 


PROPOSITION XX. THEOREM. 


The square of the semi-ordinate to any diameter, is to the rectan- 
gle under the absciss@, as the square of the semi-conjugate to 
the square of the semi-diameter. 

That is, 
IfQgq be an ordinate to any diameter CP, 
Qo e Pi tia Dt. CP 
Produce Qq to meet the major axis 
in E ; 

Draw QX, DW, perpendicular to 

the major axis, and meeting PC in X 

and W. 

Then, since triangles CPT, CoH, are 
similar, 
trian. CPT : trian. CvuE : : CP? : Cy’ 
or, trian. CPT: trap. TPvE:: ?CP*: CP?—Cr’ 
Again, since the triangles CDW, vQX, are similar, 
triangle CDW : trangle vQX :: CD? : vQ’ 


But triangle CDW = triangle CPT ; Prop. XVIII, 
or (5, 
And triangle vQX = trapez. TPvE: Prop. XVIIL, 
Cor. 3. 

ereOPpicy “CD a CRE SC irre v@? 

Or. Qu Poppa: CL? Ces CPs 


Cor. 1. The squares of the ordinates to any diameter, are 
to each other as the rectangles under their respective ab- 
Sc1Sse. 

Cor. 2. The above proposition is merely an extension of the 
property already proved in Prop. XII, with regard to the re- 
lation between ordinates to the axis and their abscisse. 


PROPOSITION XXI. THEOREM. 


The equal conjugate diameters of all ellipses described on the 
same anis major, all terminate in the same right lines. 


Let any number of ellipses 
HJ, FG be described on the same 
axis major, AB, and the right co- 
ordinates to that axis drawn from 
the extremities of the equal con- 
jugate diameters, will all coincide 
in the same lines KL, MN, or the 
vertices of al] their equal diame- 
ters will all be found in those 
lines. 

Describe on the same axis as a 
diameter, the circle AEBD, bi- 

19 


214 CONIC SECTIONS. 


sect the ares EB in K and AE in M, from which points through 
the centre, C, draw the conjugate diameters KN, ML; join 
KL, MN ; draw also the chords BE, BF, BH, also re diam- 
eters fn, hi, and qi, sr. 

The triangles ECB, KPC, being similar, their sides are pro- 
portional ; and since KC = Be and KP = BR, the triangle 
KPC = the triangle BRC = } the triangle ECB. 


Hence, CP? + PK? = 2 CB’ + + CH? 
and KO = ee EC + CB, 
or, KC? + CL? = CB? + CE? 
Hence AB? + ED? = KN’? + ML’ 


And since the semi-ordinate KP: fP: : EC : FC the tri- 
angles FCB, f PC, are similar, and fPC + 1 FCB and fC? = 

1 FB? = 1 FC?+1CB 
Hence, fC + Chi= EC? +.Be? 
Bea fv’ + qv? = FG* + AB’ agreeably to Prop. 

Ve 

Also in the triangles HCB, hPC, being for the same reason 
as before shown, similar, APC = 7 HCB, and AC? = } HB’ = 
HC’ 4 2:cbB’, and HC" -+ rC*? = HC? + CB’ 

Therefore, hi? + s7* = HE + AB? agreeably to the pro- 
perty of the ellipse. 


Cor. 1. Cpr Cb 
and 156, yee aac OA 
or, me - On 


Cor. 2. If AB, instead of being the axis major, should be 
the minor axis of a series of ellipses, the vertices of their 
equal conjugate diameters would still all be found in the lines 


MN, KL, produced. 


HY PERBOLA. 


DEFINITIONS. 


1. An Hyprrgota is a plane curve, such that, if from any 
point in the curve two straight lines be drawn to two given 
fixed points, the excess of the straight line drawn to one of the 
points above the other will always be the same. 

2. ‘The two given fixed points are called the focz. 


Thus, let QAq be an hyperbola, 8 and H the foci. 

Take any number of points in the curve, P,, P,, P,,... 

PagnePe Ps RyRy, SP SPs BF, son. tent 
HP, —SP, = HP, —SP, = HP, —SP,=.... 


g gq 


If HP, —SP, and SP’,,— HP’,...... be always equal, to 
the same constant quantity, the points P, P, P,... and P’,, 
P’,, P’,, will lie in two opposite and similar hyperbolas Q Ag, 
Q’aq', which in this case are called opposite hyperbolas. 

83. If a straight line be drawn joining the foci, and bisected, 
the point of bisection is called the centre. 

4. The distance from the centre to either focus is ealled the 
eccentricity. 

5. Any straight line drawn through the centre, and termina- 
ted by two opposite hyperbolas, is called a diameter. 

6. The points in which any diameter meets the hyperbolas 
are called the vertices of that diameter. 

7. The diameter which passes through the foci is called the 


216. | CONIC SECTIONS. 


axis major, and the points in which it meets the curves the 
principal vertices. 

8. If a straight line be drew h through the centre at right 
angles to the major axis, and witha principal vertex as centre, 
and radius equal to the eccentricity, a circle be described, cut- 
ting the straight line in two points, the distance between these 
points is called the axis minor. ~ 

Thus, let Qg, Q’q’ be two opposite Q Q 
hyperbolas, S and H the foci, join af 


MeO 

Bisect SH in C, and let SH cut the 
curves in A,a. 

Through C draw any straight line 
Pp, terminated by the curves in the 
points P, p. 

Through C draw any straight line 
at right angles to Aa, and with cen- 
tre A and radius = CS describe a circle cutting the straight 
line in the points B, b. 

Then C is the centre, CS or CH the eccentricity, Pp, is a 
diameter, P and p its vertices, Aa is the major axis, Bd is the 
minor axis. 


x 


Q’ p 
Q 
The hpyerbolas Xz, X’z', whose SE ai 


major axis is Bd, and whose minor 
axis is Aa, are called the conjugate 


hyperbolas to Qq, Q’q’. 


9. A straight line, which meets the curve in any point, but 
which, being produced both ways, does not cut it, is called a 
tangent to the curve at that point. 

10. A straight line, drawn through the centre, parallel to the 
tangent, at the vertex of any diameter, is called the conjugate 
diameter to the latter, and the two diameters are called a pair 
of conjugate diameters. 

The vertices of the conjugate diameter are its intersections 
with the conjugate hyperbolas. 

11. Any straight line drawn parallel to the tangent at the 
vertex of any diameter, and terminated both ways by the 
curve, is called an ordinate to that diameter. 

12. The segments into which any diameter produced is di- 


HYPERBOLA. © 217 


vided by one of its own ordinates and its vertices, are called 
the abscisse of the diameter. 

13. The ordinate to any diameter, which passes through 
the focus, is called the parameter of that diameter. 

Thus, let Pp be any diameter, . R 
and Té a tangent at P ; | 

Draw the diameter Dd parallel 
to Tt; 

Take any point Q in the curve, 
draw Qq parallel to T¢ and cutting 
Pp produced in v ; 

Through 8 draw Rr parallel to 
Ti; 

Then Dd is the conjugate diame- 
ter to Pp, 

Qg¢ is the ordinate to the diameter Pp corresponding to the 
point Q. 

Pv, vp, are the abscissee of the diameter Pp corresponding 
to the point Q. 

Rr is the parameter of the diameter Pp. 

14. Any straight line drawn from any point in the curve at 
right angles to the major axis produced, and terminated both 
ways by the curve, is called an ordinate to the azis. 

15. The segments into which the major axis produced is di- 
vided by one of its own ordinates and its vertices, are called 
the absciss@ of the axis. 

16. The ordinate to the axis which passes through the fo- 
cus, 1s called the principal parameter or latus rectum. 

(It will be proved in Prop. IV, that the tangents at the prin- 
cipal vertices are perpendicular to the major axis; hence de- 
finitions 14, 15, 16, are in reality included in the three which 
immediately precede them.) 

17. If a tangent be drawn at the extremity of the latus rec- 
tum, and produced to meet the major axis; and if a straight 
line be drawn through the point of intersection, at right angles 
to the major axis; the tangent is called the focal tangent, and 
the straight line the directriz. 

~Thus, form P, any point in the 
curve, draw P Mp perpendicular to Aa, 
cutting Aa in M; 

Through S draw L/ perpendicular 
to Aa; 

Let LT, a tangent at L, cut Adin T ; 

Through T draw Nn perpendicular fs 
to Aa: 

Then, Pp is the ordinate to the axis 
corresponding to the point P. 


19* 


218 CONIC SECTIONS. 


AM, Ma, are the abscisse of the axis corresponding to the 
point P, 
Li is the latus rectum, 
LT is the focal tangent, 
Nn is the directrix. 

18. An asymptote is a diameter which approaches the curve 
continually as they are both produced, but which, though ever 
so far produced, never meets it. 

19. If the asymptotes of four opposite hyperbolas cross 
each other at right angles, the hyperbolas are called right 
angled or equilaterial hyperbolas. 


PROPOSITION I. THEOREM. 


The difference of two straight lines drawn from the foci to any 
point in the curve, is equal to the major axis. 


That is, if P be any point in the curve, F 
HP—SP = Ad: 


For, 
HP-SP=AH-AS=Aa+aH-AS 

And, Def. 1. 
HP-SP=aS-aH=Aa-aH+AS 

Or, 
2(HP — SP) = 2Aa 

HP —SP = Aa 


Cor. 1. The centre bisects the major axis ; for, since 
AH — AS = aS — 


Or, SH — 2AS = SH — 2aH 
Bs AS = aH 
And CS = CH, by def. 3. 
AC = aC. 
Cor. 2. 


2AC + 28P. 


PROPOSITION Il. THEOREM. 
The centre bisects all diameters. 


Take any point P in the curve; 
Times. Ee sckh,- bs 9, La 
Complete the par allelogram SPHp ; 
join C,p; 0, P; 
Then, since the opposite sides ofpa- H 
rallelograms are equal, 
HP = Sp, SP Hp} 
«. HP —SP=Sp — Hp; 


- 
HYPERBOLA. 219 


.". p is a pointin the opposite hyperbola by definition Q. 

Again, since the diagonals of a parallelogram bisect each 
other, and since SH is bisected in C, (def. 3,) 

.. Pp is a straight line and adiameter, and is bisected in C. 

In like manner, it may be proved that any other diameter is 
bisected in C. 


PROPOSITION III. THEOREM. 


The rectangle under the segments of the major axis produced, 
made by the focus and its vertices, is equal to the square of 
the semi-axis, minor. : 


That is, 
AS .Sa 
For, 


BC? 
BC? = AB? — AC’ 
SC? — AC’, by def. 8, 
(SC — AC) (CS+AC) 
AS . 8a 
Cor. The square ef the eccentricity is 
equal to the sum of the squares of the 
semi-axes. 
For, SC? =”AB*, def. 8, 
= AC’ + BC’. 


PROPOSITION IV. PROBLEM. 
To draw a tangent to the hyperbola at any point. 


Let P be the given point ; 

Join S, P.; H, P; 

Bisect the angle SPH by the 
straight line T?. 

Ttis a tangent to the curve at P. 

For if Tt be not a tangent, let Té 
cut the curve in some other point p. 

Joia 8S, p ; H, p; draw SYO per- 
pendicular to Tt, meeting HP inO ; 
join p, O. 

Since the angles at Y are right angles, and angle SPY 
= angle OPY by construction, and side YP common to the 
two triangles SYP, OYP. 

o SY = OY 
ABO 5 PaysetOlP: 

Again, since SY = OY, and Yp common to the two trian. 

gles SYp, OYp, and the angles at Y equal ; 
o's Sp = Op { 


ey 


220 CONIC SECTIONS. 
1. Hp—Op= Hp—Sp 
— HIPp_sp 
— HP_opP 
— HO 


it Hp =#H0;+.0p 
that is, one side of the triangle HOp is equal to the other two, 
which is absurd ; 
.. pis nota point in the curve: and in the same manner, it 
may be proved that no point in the straight line T¢ can be in 
the curve, except P ; 
, .. Tt isa tangent to the curve at P. 


Cor. 1. Hence tangents at A and a, are perpendicular to 
the major axis. 
Cor, 2. SP and HP make equal angles with every tangent. 
Cor. 3. Since SPH, the vertical angle of the trian. SPH, is: 
bisected by the straight line PT, which cuts the base in T, 
ett ee are OL 


PROPOSITION V. THEOREM. 
Tangents drawn at the vertices of a diameter are parallel. 


Let Tt, Ww, be tangents at P, p, 
the vertices of the diameter PCB. 
IOI Py, bu Liye uaa ect. 
Then, by Prop. Il, SH isa paral- 
lelogram, and since the opposite an- 
gles of parallelograms are equal, 
*. angle SPH = angle SpH. 
But the tangents Tt, Ww; bisect 
the angles SPH, SpH, respectively. y 
*, angle WpS = angle HPT 
= angle PTS, which is the exterior oppo- 
site angle to WpS. 


.. Ww is parallel to Te. 
. ft 

Cor. If Dd be a diameter conjugate 
to Pp, and terminated by the conju- 
gate hyperbolas, tangents drawn at D 
and d will be parallel. 

Hence tangents drawn at the extre- 
mities of conjugate diameters form a 
parallelogram. 


_ -HYPERBOLA. 221 

! | PROPOSITION VI. THEOREM. " 

If straight lines be drawn from the foci to a vertex of a diame- 

ter, the distance from the vertex to the intersection of the con- 

jugate diameter with either focal distance, is equal to the 
semi-axis mujor. 


That is, if Dd be a diameter to conju- 
gate to Pp, cutting SP produced in EH, 
and HP in e, 

Por 'Pe = AC. 

Draw HI parallel to Dd, meeting SP 
produced in [. 

The angle PHI = alternate angle 
He 'E 

= angle TPS 
= angle HIP 

‘.* HI is parallel to Dd or Tt 

Pik Te 

Also, since SC = HC, and CEH is parallel to HI, the base of 
the trian. SHI, 

.. SE = KEI. Hence, -.. PE = PI—EI = HP—SE=HP 
SP—PE 

“2PE= HP —SP = 2AC 

PEs AC: 

Also angle PEe = angle PeE, ... Pe = PE and 

bene. 


PROPOSITION VII. THEOREM. 


Perpendiculars from the foci upon the tangent at any point, in- 
tersect the tangent in the circumference of a circle whose dia- 
meter is the major aats. 


From 8 let fall SY perpendicular on 
Tt a tangent at P. 
Join S, P.; H, P ; let HP meet SY 
ings siomm Ce Y 5 
Then, since angle SPY = angle 
KPY, and the angles at Y are right 
angles, and the side P Y common to the 
two triangles SPY, KPY, 
See a ag 
and KP*=SP. 
Again, since SY = YK, and SC = CH, CY cuts the sides 
of the trian. HSK proportionally, 
.. CY is parallel to HP. 


222 CONIC SECTIONS. 


Also, since CY is parallel to HP,SY = KY, and SC = CH, 
A =iHK =; (HP—KP) = ! (HP—SP) =} 
AC. 

Hence, a circle described with centre C and radius = CA, 
will pass through Y, and in like manner, if HZ be drawn per- 
pendicular to Tt, it may be proved that the same circle will 
pass through Z also. 


PROPOSITION VIII. THEOREM. 


The rectangle contained by perpendiculars from the foci upon 
the tangent at any point, is equal to the square of the semi- 
axis, minor. 

That is, 

Sethi 7a) BCS ‘ 
Let Tt bea tangent at any point P; 
On Aa describe a circle cutting T¢ in 

Weanid 2.2 40ny Ney 551," 

Then, by last Prop. SY, HZ are 
perpendicular to Té. 

Let HZ meet the circumference in z ; 

Join (230, Ys 

Since zZY is a right angle, the seg- 
ment in which it lies is a semicircle, 
and z, Y,are the extremities ofa diameter; 

. .. 2CY is a straight line and a diameter. 
Hence the triangles CXS, C2H are in every respect equal 


Pept t= He 
gy a7 . HZ 
SIT A ~ Ha. 


= BC?. Prop. IIL. 
PROPOSITION IX. THEOREM. 


Perpendiculars let fall fromthe foci upon the tangent at any 
point, are toeach as the focal distance of the point of con- 
tact. 


That is, 
Bren A ors bo: Epes 
For the triangles SPY, HPZ, are 
manifestly similar ; 
ei pe lin sO P PAP: 
Cor. Hence, 


SP 
SP 


Bo. me last Prop. 


i 


ke ee ee ee 


HYPERBOLA. 223 
SP 
any 2 AC 4 S8P 
So also, : 
, ; 1E ere . HP 
HZ? = BC SP 8 o1 & peer 


PROPOSITION X. THEOREM. 


If a tangent be applied at any point, and from the same point an 
ordinate to the axis be drawn, the semi-axis major 1s a mean 
proportional between the distance from the centre, to the or- 
dinate with the axis, and the distance from the centre to the 
intersection of the tangent with the axis. 


That is, 

CT :CA::CA:CM. 

Since the angle SPH is bisected by 
PT, which cuts HS, the base of the 
triangle HPS, in T, «. 

HT : ST eee Pies Sr 
~. HT —ST:HT+ST:: HP — SP: 


OF, fs) 2 GLa 2: ZAC . sHbP 
P 


202 AC -: SHY HHP -SPiiae en eee} 
But since PM is drawn from the vertex of triangle HPS 
perpendicular to HS produced, 

HM—SM:HP+SP::HP—SP:HM+SM 
of, “SH THP+SPS:) 2.4C 2. QuCMy- - 9" ahh oes 
Comparing this with the proportion marked (1), we have 

Oe ri eee AGF Wr une A oe se 2G 

Ore: CAs CA sé CM. 


PROPOSITION XI. THEOREM. 


Tet AQa be a circle described on the major axis, from the point 
T,draw TQ perpendicular to Aa, meeting the circle in Q, 
join QM. : 

Then QM is a tangent to the circle 
at Q 

Join C, Q. 

For if QM be not a tangent, draw 
QM’ a tangent at Q, cutting AC in 
M’ 

Then CQM’ is aright angle. 

“. Since QT is drawn from the. 
right angle CQM’ perpendicular to 
the hypothenuse, 


poke 2 CONIC SECTIONS. 


im = COs sown CT. 
war, CMW? CA 7:28CAr OT). CQ = CA. 
But by the last Prop... | ° 
CM: UA *#@CA.: CT 
. CM = CM’, camel 
which is absurd; .*. QM is nota tangent at Q; and in the 
same manner it may be proved that no line but QM’ canbe a 
tangent at Q. : 


PROPOSITION XII. THEOREM. 


The square of any semi-ordinate to the axis, is to the rectangle 
under the absciss@, as the square of the semi-axis minor, is 
to the square of the semi-axis major. 


Thatis, if P be any point in the 

curve, 
PM’ aM. Ma@:: BC?: AC’. 

Describe a circle on Aa, and draw 
PT a tangent to the hyperbola at P, 
intersecting the circle in the points 
Y, Z, and the major axis in T. 

‘Draw TQ perpendicular to Aa, 
meeting the circle inQ; jon QM 

Then QM is a tangent to the circle at Q by Prop. II , and 

*, the angle CQM is a right angle. 

Jon 8, Y; ,Z; HSW and HZ are perpendicular to Tt, 
Prop. VIL. 

Hence the triangles PMT, SYT, PZT, are similar to each 
other. 

SOR MENS. ee I eee OLY 
and, PM : HZ sw MT. TZ 

2uPM*: SYS eeMT os WD Yor TZ) 

“vin ah GAA Relea to | @Rainhaaaeae gets Mi bye aarp ¢ hes 


Prop. VIL. 
QM’: CQ’, 
~ MQT. MCQ are cee triangles, 
>: AM.Ma: AC’, 
-PM?:AM.Ma:: BC? : AC? 
Cos lL " 
Let P, M,, P, M,, ...... be ordinates to the axis from any 
point P, Py Sass 
Then by Prop. 


Preise a... Ma: : BC? : AC? 
P,M,* : AMj@M,a:: BC? ; AC’ 
Dis Py aan, tes A ee M,a: AM, .M,a, 


HYPERBOLA. 225 


That is, the square of the ordinates to the axis are to each 
other as the rectangles of their abscisa. 
Cor. 2. By Prop. | | 
PM?: AM.Ma_ :: BC? : AC 
But AM = CM—CA, Ma = CM+CA, 
ee PVA? CMC A? ih BEF =: AC?. 


CR? : CA?.: CHR+-Cad? : dP’, 
since dP? = CD’. 


Cor. 3. Since by the proposition kat Zp 
PM: AM: Ma:: BC?: AC’, te | 
we have in the conjugate hyberbolas #\ | 6 i 


ow 


PROPOSITION XIII. THEOREM. 


The latus rectum is a third proportional to the axis major and 
minor. 


That is, 
Aa > Bb-: : Bo:; Li. 
Since LS is a semiordinate to the 


axis. \ 
AC? : BC?: : AS ..S@: Ls’; 
Prop. XII. — 
BC* ef C873 
Prop. II. os 


RAC BCS BO MES 
or, Aad . Bb. 3: Bb? > Li. 


PROPOSITION XIV. THEOREM. 


The area of all parallelograms, formed by drawing tangents at 
the extremities of two conjugate diameters, is constant, each 
being equal to the rectangle under the axes. 


Let Pp, Dd, be any two conjugate 
diameters, WwXz, a parallelogram 
inscribed between the opposite and 
conjugate hyperbolas by drawing 
tangents at P,p, D,d; then Pp, Dd, 
divide the parallelogram WzXw 
into four equal parallelograms. 

Draw Pm. dm, ordinates to the 
axis ; PF perpendicular to Dd. 

Let CA meet PX in T and Wz int; 

Then Cis C2 32. GA- 3M 

20 


226 CONIC SECTIONS. 


Ct : CA :: CA: Cm, Prop XI: 
OT 2 Ops: Om: CM 
But CT : Ct :: MT: Cm, by similar triangles. 
MT: : Cm 221 0m:: CM 
. CM. MT = Cm? - = = = = = = (1) 


Again,” CM: CA 23:GA : CT 
« CM: CA :: MA: AT,dividendo : 
Or, “GMs MateeMac: MT, componendo : 
pa a CM. MT) Gale ae ee (2) 


But, AC?: BC?:: AM: Ma: PM? 
Or, ACG’ BU: Cn? : PM 
igh ACe:?BC 2: Cm > PM 
Similarly, AC : BC :: CM : dm 
Or, -BG.: dmiz: CA: CM 
But, CT xs CAs: CA]: (CM 
$e OT CA 2 BCs dm 
But, PE : CT?:% dm .Cd 
.. PRGA. BCS 3G 
.. Rectangle PF .CD = rectangle AC . BC 


or, Parallelogram CX = rectangle AC . BC 
*. Parallelogram WwXz = 4 AC. BC=Aca. Bb 


Cor. By (2), 
Cm = HM. Ma 
ces (CM—CA) (CM+CA). 
= CM’—CA 
CA? = CM’—Cm? 


yen similarly, CB’ = dm’—PM?. 
PROPOSITION XV. THEOREM. 


The difference of the squares of any two conjugate diameters, is 
equal to the same constant quantity, namely, the difference of 
the squares of the two azes. 


That is, if Pp, Dd, be any two con- 

jugate diameters, 
Pp?— Dd* = Aa’* — Bb’. 

Draw PM dm, ordinates to the 
axis. 

Then, by Cor. to last Prop. 
AC?—BC*=CM?+ PM?--(Cm*+d’m) 

=CP’—Cd@’ 

Ad—BbL’ =Pp*—Da’. 


HYPERBOLA. 227 


PROPOSITION XVI. THEOREM. 


The rectangle under the focal distance of any point, is equal to 
the square of the semi-conjugate. 


That is, if CD be conjugate to CP, 

ey ees ke by =3, OD)": 

Draw SY, HZ, perpendiculars to 
the tangent at P, and PF perpendicu- 
lar to CD; 

Then by similar trian. SPY, PEF 

Peete Mada eb e PE 
OF be ae vas we DOs PE 
“EN = ACY by-Prop. VI. 

simslarly; HPs sy Hee CAC) ss PE 
Pape SY As ACs Ph 
CO DFs CO Eby Prope Lys 

But SY HAC Be by Praps vill 

Peerks vekb bes OL). 


PROPOSITION XVII. THEOREM. 


If two tangents be drawn, one at the principal vertex, the other 
at the vertex of any other diameter, each meeting the other’s 
diameter produced, the two tangential triangles thus formed 
will be equal 


That j is, 

triangle CPT = triangle CAK. 

Draw the ordinate PM ; then 

OVE AGA 22 Ch: CK, by similar 

triangles. a 
But, CM :. CA ::CA:CT 

a OF OE MEET 6) ma (Oa 

The two triangles CPT, CAK, have 
thus the angle C common and the 
sides about that angle reciprocally 
proportional ; these triangles are .*. equal. 


Cor. 1. Take each of the equal triangle CPT, CAK, from 
the common space CAOP ; there remains 
triangle OAT = OKP. 


Cor. 2. Also take the equal triangles CPT, CAK, from the 
common triangle CPM; there remains 


triangle MPT = trapez. AKP M. 


228 CONIC SECTIONS. 


PROPOSITION XVIII. THEOREM. 


The same being supposed, asin last proposition, then any 
straight lines QG, QE, drawn parallel to the two tangents 
shall cut off equal spaces. 


That is, 
triangle GQE = trapez. AKXG 
triangle rgH = trapez. AKRr 

Draw the ordinate PM. 

The three similar triangles CAK, 
CMP, CGX, are to each other as c 
CAI CMA CGs 
~. AKPM: trap. AKXG : : CM’—CA? : CG’—C A®, dividendo. 
But, PN. QG* :: CM’#CA?’ : CG’—CA’, 

.. trap. AKPM : trap. AK XG: : PM eet 
But, trian. MPT : trian. GQF:: PMia? Qe 
*.* the triangles are similar. 

..trap. AKPM: trian. MPT: : trap. AK XG: trian. GQE, 

But, by Prop. X VII, Cor. 2. 

trap. AKPM = triangle MPT ; 
.. trap. AKXG = triangle GQE. 
And similarly,trap. AKRr = triangle rgE. 


Cor. 1. The three spaces AKXG, TPXG, GQK, are all 


equal. 


Cor. 2. From the equals, AKXG, EQG, take the equals 
AKRr, Egr ; there remains, 
RrXG, = rqgQG. 


Cor. 3. From the equals RrXG,rqQG, take the common 
space rqgvXG ; there remains, 
triangle vQX = triangle vgR. 


Cor. 4. From the equals EQG, TPXG, take the common 
space EvXG ; there remains, 


TPvE = triangle vQX. 


Cor. 5. If we take the particular 
case in which QG coincides with the 
minor axis, . 

The triangle EQG becomes the tri- 
angle IBC, 

The figure AKXG becomes the tri- 
angle AKC, 

.. triangle IBC = triangle AKC 
= triangle CPT. 


HYPERBOLA. 229 
PROPOSITION XIX. THEOREM. 


Any diameter bisects all its own ordinates. 
That is, Q 
If Qgq be any ordinate to a dia- 
meter CP, 
Ui ug 
Draw QX, 2, at right angles to 
the major axis ; 
Then triangle vQX = trian- 
gle vgx ; Prop. XVIIL, Cor. 3. 
But these triangles are also 
equiangular ; 


Qu =! 19: 


Cor. Hence, any diameter divides the hyperbola into two 
equal parts. 


PROPOSITION XX. THEOREM. 


The square of the semi-ordinate to any diameter, is to the rec- 
tangle under the abscisse, as the square of the semi-conjugate 
to the square of the semi-diameter. 


That is, 
If Qq be an ordinate to any dia- 
meter CP, 
net: Ror inane a Gre 
Let Qq meet the major axis in E; 
Draw QX, DW, perpendicular to 
the major axis, and meeting PC in 
X and W 
Then, since the triangles CPT, 
Cv, are similar, 
trian. CPT: trian. CuK :: CP? Cv* 
or, triad, GPU scirap. PPK >: OR: 
Cv? —CP” 
Again, since the triangles CDW, vQX, are similar, 
‘triangle CDW : triangle DEL ME A OD ts at): 
But, triangle CDW = triangle CPT ; Prop. XVIII, Cor. 5, 
And triangle Ree = tr apr TPE; Prop. XVIIL, Cor. 3. 
GRS asCNye Cv! CP? : vQ? 
Or, Qv? :Pu.vp: OP oy Phi MOM 
Cor. 1. The squares of the ordinates to any diameter, are 
to each other as the rectangles under their respective abscisse, 


Cor. 2. The above proposition is merely an extension of 
the property already proved in Prop. XII, with regard to the 
relation between ordinates to the axis and their abscisse. 


230 CONIC SECTIONS. 
PROPOSITION XXI. THEOREM. 


If tangents be drawn at the vertices of the axes, the diagonals 
of the rectangle so formed are asymptotes to the four curves. 


Let MP meet CE in Q ; 
Then, MQ?: CM?:: AE? : AC’ 
22 BOA 
>: MP? :CM?—CA”. . 
Now, as CM increases, the ratio of CM? 
to CM’—CA’* continually approaches to a 
ratio of equality ; but CM* — CA’ can never 
become actually equal to CM’, however 
much CM may be increased. Hence, MP 
is always less than MQ, but approaches con- 
tinually nearer to an equality with it. 
In the same manner it may be proved, that CQ is an asymp- 
tote to the conjugate hyperbola BP’. 
Cor. 1. The two asymptotes make equal angles with the 
axis major and with the axis minor. 
Cor. 2. The line AB joining the vertices of the conjugate 
axes is bisected by one asymptote and is parallel to the other. 
Cor. 3. All lines perpendicular to either axis and is termi- 
nated by the asymptotes are bisected by the axis. 


PROPOSITION XXII. THEOREM. 


If a line be drawn through any point of the curves, parallel to 
either of the axes, and terminated at the asymptotes, the 
rectangle of its segments, measured from that point, will be 
equal to the square of the semi-axis to which it is parallel. 

That is, 


SS 
the rect. HEK or HeK = CA’, 


and rect. AEK or hek = CA’. ZB 


For, draw AL parallel to Ca, and aL to CA. Then 
by the parallels. CA? :C or AL? :: CD’: DH’; 
and by Prop. XI, CA’: Ca? ::CD*—CA?: DE’ ; 
seby subpreo As Cate: CA? : DH" Et Ors Re 
But the antecedents CA’, CA’ are equal, 


} 
d 
Fr 


oo 


HYPERBOLA. 231 


Therefore, the consequents Ca’, HEK must also be equal. 
In like manner it is again, 
by the parallels, CA’ : Ca’ or AL’: : CD? : DH’; 
CA’? :Ca:: CD?+CA?: De’; 

“. by subtr. CA? : Ca? : : CA* : De’-——DH? or HeK. 
But the antecedents CA’, CA’ are the same, 
.. the conseq. Ca’, HelX must be equal. 
In like manner, by changing the axes, is Ak or hek = CA’. 


Cor. 1. Because the rectangle HEK = the rectangle HeK, 
te Ee Feb: Beh. a 
And consequently HE : is always greater than He. 


Cor. 2. The rectangle KEK = the rectangle HEA. 
For, by similar triangles EA : EH :: Ek: EK. 


Scholium. It is evident that this proposition is general for 
any line oblique to the axis also, namely, that the rectangle of 
the segment of any line, cut by the curve, and terminated by 
the asymptotes, is equal to the square of the semi-diameter to 
which the line is parallel—since the demonstration is drawn 
from properties that are common to all diameters. 


Cor. 3. Hence it is evident that all the rectangles are equal 
which are made of the segments of any parallel lines, cut by 
the curves, and limited by the asymptotes ; and therefore, that 
the rectangle of any two lines drawn from any point in the 
curve, parallel to two given lines, and limited by the asymp- 
totes, is a constant quantity. 


GENERAL REMARK. 


Having thus discussed at length, the properties of the para- 
bola, ellipse and hyperbola, in the relations of their local and 
peculiar constructions, it may be observed, that there are many 
properties common to each—especially in the ellipse and 
hyperbola. These curves have many striking similarities in 
their determinations, although there is but little similarity in 
their construction ; for the axes, of the hyperbola are thrown 
without the curve, while in the ellipse they are within ; hence it 
should not be surprising, that the same or corresponding 
quantities, thus differently associated, should propagate by a 
somewhat similar condition of their mutations, curves so ap- 


232 CONIC SECTIONS. 


parently dissimilar in their developments. The ellipse is a 
curve of limited extent returning into itself as the circle, but 
the hyperbola is unlimited in its construction, or its determin- 
ation, for its branches may be extended indefinitely ; the same 
may be observed in relation to the parabola, which may be 
indefinitely extended, and the branches of the curve become 
at length parallel to its axis ; this, however, is only in their in- 
finite extension. But the hyperbolic curve never approaches 
toward or evento a parallelism with the axis, but approaches 
infinitely toward its asymptotes; but without ever touching 
them, except in their infinite extension. The manner in 
which these curves are derived from the sections of a cone, as 
their name indicates, their origin will be shown in another 
volume, and their quadratures, and some other properties in 
relation to them, will be there discussed. 


HIGHER GEOMETRY 


MENG SU RAST ol Or N: 


FOURTH PART OF A SERIES 
ON 


ELEMENTARY AND HIGHER 


GEOMETRY, TRIGONOMETRY, AND MENSURATION, 


CONTAINING MANY VALUABLE DISCOVERIES AND IMPROVEMENTS IN MATHEMATICAL 
SCIENCE, ESPECIALLY IN RELATION TO THE QUADRATURE OF THE CIRCLE, 
AND SOME OTHER CURVES, AS WELL AS THE CUBATURE OF CERTAIN 
CURVILINEAR SOLIDS 5 DESIGNED AS A TEXT-BOOK FOR COLLEGIATE 
AND ACADEMIC INSTRUCTION, AND AS A PRACTICAL 
COMPENDIUM OF MENSURATION. 


BY NATHAN SCHOLFIELD. 


NEW YORK: 
PUBLISHED BY COLLINS, BROTHER & CO. 


No. 254 Pearl Street. 


1845. 


Entered according to Act of Congress, in the year 1845, by 
_ NATHAN SCHOLFIELD, 
In the Clerk’s Office of the District Court of Connecticut. 


G. W. WOOD, PRINTER, 29 GOLD ST., NEW YORK. 
——_— —  ————————— 


PREFACE. 


Havine, in the former parts of this series, treated of the 
elements of geometry, trigonometry, conic sections, &c., it 
now remains for us, in accordance with our original design, 
to make such application of the former principles, as to elicit 
such other truths or principles as depend on their various 
combinations, and to investigate the relations of such subjects 
as pertain to the higher geometry. And, without attempting 
to give a full and perfect treatise on the subject, which would 
require volumes, we shall endeavor to present some portions 
of this ancient subject in a new dress; hoping, thereby, to 
render its beauties more plainly visible, and its oracles more 
intelligible. 

Some new solids are introduced into this volume, the most 
important of which is a class termed revoloids ; which, from 
their organization, seem to serve as a connecting link between 
rectilinear and curvelinear solids. The properties of those 
solids are discussed, and their surfaces and solidities are de- 
termined. Some new curves are also introduced and inves- 
tigated, among which is the revoloidal curve, whose quadrature 
is determined; and, from its relation to the circle, and also to 
rectilinear figures, we are enabled to approximate to the cir- 
cle’s quadrature to an indefinite extent. During the investiga- 
tion of this subject, other important properties of the circle 
will be developed, by which the area of the segment of a 
circle whose are and sine are known, may be computed with 
as little labor as that of the area of a triangle whose base 
and perpendicular are given. 

We have also introduced into this work a mode of con- 


lV PREFACE. 


struction for variable quantities, or such magnitudes as depend 
on variable factors ; and have adapted a notation, embracing 
some of the principles of the calculus, by which variable 
magnitudes may be algebraically discussed, and their condi- 
tions rendered intelligible. By this notation, some of the more 
difficult geometrical subjects are susceptible of the most ele- 
gant solution; and we are also enabled to get a definite 
algebraic expression for the circle’s quadrature, in terms of 
the diameter. The mensuration of such superficies and solids 
as depend on the higher geometry, follows at the close of 
the work. 

From the hasty manner with which a considerable portion 
of this work has been prepared, it can hardly be presumed 
to be entirely free from errors; but it is believed that if any 
errors exist, they are such as involve no important principle. 

The author, with these remarks, submits the work to the 
consideration of an intelligent public. 


CONTENTS. 


BOOK I. 


PAGE. 


On the Species and Quadrature of the Sections of Elementary Saga 
embracing the Oa bag rath and Gab etae 

Definitions, 

The Sections of a Cone, : 

The Sections of a Polyedroid, 

The Sections of a Prism, &c., . 

Quadrature of the Parabola, ; 

On the Ellipse, its Quadrature, &c., 

On the Hyperbola, its Quadrature, &c., 

Equations to the Conic Sections, 


BOOK II. 


Solid Sections, or Segments of Solids of Revolution, Ungulas, &c., 
Definitions, ; ; : ; ; i 
Cylindrical Ungulas and Segments, ; 

Conical Segments and -Ungulas, ‘ 

Segments and Ungulas of an Elliptical Cylinder, 

Spherical Ungulas and Segments, 

Parabolic Prisms and Ungulas, 


BOOK III. 


On Revoloids and Solids formed by the Revolution of the Conic Sra a 
Definitions, : : 
Quadrature of the Surface of a avalold determined, 

Revolvoids and Ungulas equivalent to a Sphere or Spheroid, 

‘The Cubature of the Revolvoid determined, ; : 

Segments of Revolvoid, 

Parabolic Revoloid, its Cubature determined, ‘ 

Sections of Solids formed by the Revolution of the Conic Sections, 
Scholia and Formule, in relation to Cylindric and Conical Ungulas, . 


BOOK IV. 


On the Revolvoidal Curve, the Rectification of the a and other 
Curves, and on the Quadrature of the Circle, ‘ 

Definitions, : 

Projection of a Revolvoidal Curve, 

Equation to the Curve, 

The Ellipse, its Rectification, &e. ee 

Rectification of the Revoloidal Curve, 

On the Circle’s Circumference, and its Quadrature, 

Curve of the Circle’s Quadrature, . 

On the Quadrature of the Segment of a Circle, 

On Spirals—their eae so OLB. : 5 

On the Cycloid, 


vi CONTENTS. 


BOOK V. 


On the Production and Resolution of Geometrical Magnitudes, . 
Cuap. I[.—Definitions and Principles, . ; : 
Production of Surfaces and Solids, 


Cuap. I].—On the Construction of Quantities Pevhose Hlerments are a 
ee Poe | 
aoe 
. 159 


series either of constant or variable quantities, 
Explanation of Principles, and Notation, 
Construction of Variables, 
Construction of Curves from their Equations, | 


Quadrature of the Circle expressed algebraically, in terms 


of known functions of the diameter, 


-3 
Equivalent Constructions for the Equations of Curvelinear 
: : at OL 
. 166 
. 166 


Solids, : : 
Cuar. Il.—The Differential and Integral Calculus, 4 
Differential Calculus, . 

Integral Calculus, 


Cuar, 1TV,—On the Centre of Sureche aril Solids, the Virtual Centre 
AS 
aye br gs} 
a hey! 
175 
2179 
. 180 
. 181 
Cuap. V.—On the Relations of Lines, Surfaces, and Solids, gene- 
. 182, 
182 


and Centre of cere OO cae 
Definitions, 
Virtual Centre of a Sy stem of Points, 
Virtual Centre of Surfaces, 
The Virtual Centre of Solids, 
The Virtual Centre of a Circular Arc, 
The Virtual Centre of the Surface of a Solid, 


rated by the motion of the Virtual Centre, . 
General Proposition, ; : 


MENSURATION OF SUPERFICIES. 


Circular Segments and Zones, . : 

To find the Circumference of a Circle, or any Are, 

Mensuration of the Ellipse, : ; 

Mensuration of the Parabola, 

Mensuration of the Hyperbola, . 

To find the Area of any Plane Surface by ‘Equi-distant Ordinates, 


MENSURATION OF SOLIDS. 


Mensuration of the Sphere and Revoloid, 

Segments of a Sphere or Revoloid, 

Segments of a Spheroid, . 

Mensuration of the Paraboloid, &c. opie : : 5 : 
Mensuration of the Hyperboloid, Sens : ; ‘ : ; 
Circular Elliptical and Parabolic Petey : - : 
Ungulas, . 

Rings, . 

Gauging of Casks, 


SPECIFIC GRAVITY OF BODIES. 


Table of Specific Gravities, 
Magnitude of Bodies, by their Weight and Specific Gravity, 


QUESTIONS FOR EXERCISE, 


Description of an Instrument for Measuring Distances and Heights by 


a single observation, 
Notes, y 


PAGE. 
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159 
160 


169 


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Bib yy 
IoD 
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. 207 
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. 214 
. 216 
. 218 
ey bs) 
. 221 
. 224 
. 227 
. 228 


. 233 
. 234 


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. 242, 
. 248 


HIGHER GEOMETRY. 


PART II.-—-BOOK I. 


SPECIES AND QUADRATURE OF SUPERFICIAL SECTIONS OF 
ELEMENTARY SOLIDS. 


DEFINITIONS. 


1. Superficial sections are surfaces formed when solids are 
cut by plane or curved surfaces. 


2. If the cutting surface is a plane, the section is a plane 
seclion. 


3. Superficial sections of solids take different names, ac- 
cording to the form of the solid in the plane of the section. 


4. From the cylinder, we have the rectangle, the circle, 
and, as will be shown, (Prop. VIII. Cor.) the edlipse. 


5. From the cone, we have five different figures, viz: a tri- 
angle, a circle, and, as will be shown in Propositions I, IL, 
and III., a parabola, an ellipse, and a hyperbola. 


6. From the sphere, we have only the circle. 


Scholium. The parabola, ellipse, and hyperbola, will be 
more specially the subjects of this book. 


8 SPECIES AND QUADRATURE OF 
PROPOSITION I. THEOREM. 


If aright cone BEG be cut by a plane Ap o, which is parallel 
to a plane touching the cone along the slant side BH, the sec- 
tion A po is a parabola. 


Let BEG be that position of the 
generating triangle which is perpen- 
dicular to the cutting plane Apo; 
An their common section, which is 
parallel to BE. Then, since the 
plane BEG passes through the axis, 
it is perpendicular to the base KoG, 
and to every circular section CPD 
parallel to the base; it is also per- 
pendicular to Apo. Hence, the 
common section PO of the planes 
A po, CPD, is perpendicular to BEG 
and therefore to An and CD. 

But, AN : ND: : BE: EG, which is_a constant ratio ; 
therefore, by the properties of the circle. AN: ND::CNx 
ND: NO’, since CN is equal and parallel to En, and constant. 
Hence the curve is a parabola whose axis is An. 

Cor. If L be the latus rectum of the parabola pAo, Lx AN 
NE a—GN XIND: 


PROPOSITION II. THEOREM. 


If a cone BEG be cut by a plane KAP through both slant sides 


the section is an ellipse. 


Let BEG be that position of the 
generating triangle which is perpen- 
dicular to the cutting plane: CPD 
any circular section. Draw AHK 
parallel to EG, and therefore bisect- 
ed by the axis BO. 


Then EN: CN:: EA: AK 
NA:ND:: EA: EG; | 
ENXxXNA :CNXND (NP’) 
-: FA’?: EGxX AK 


which is the property of an ellipse, 
one of whose axes is EA and the 
other a mean proportional between EG and AK. (Conic Sec- 
tions, Ellipse, Prop. XI) 


SECTIONS OF ELEMENTARY SOLIDS. 9 
PROPOSITION II. THEOREM. 
If a right cone BED be cut through one side BE by a plane 


RAP which being produced backwards, cuts the other side 
DB produced, the section is an hyperbola. 


M 


Let DGEH be any circular section, BGH a triangular sec- 
tion through the vertex B of the cone parallel to the plane 


* Then, AN: ENS: BE: EF 
; NM: ND:: BF: FD 
ANXNM: ENXND (NP"*):: BF’: EFXFD (FH?) 
which is the property of an hyperbola, whose axis major is 
AM, and whose conjugate axis is to AM as FH to BF. 


Cor. If GT, HT, be tangents to the circle at G, H; and 
planes passing through GT, HT, respectively, touch the cone 
along the lines BG, BH; also, if TB, the common section of 
the planes, meet AM in C, then the common section CO, CQ, 
of the plane RAP, extended to meet the tangent planes, are 
the asymptotes of the hyperbola. 

Draw BL parallel to DE, meeting AM in L: then the axes 
of the hyperbola being in the proportion of BF to FH, the an- 
gle GBH, or the equal angle OCQ is the angle between the 
asymptotes. 

Now, by similar triangles ALB, BFE, and CLB, BFT ; 
AL:CL:: TF: FE, and therefore AC: CL::TE:FE. In 
like manner, by similar triangles MLB, BFD, and CLB, BFT ; 
ML : CL: : TF: DF, and therefore CM: CL: : TD: DF. 
But, by the property of the circle. TE: FE: : TD: DF. 

2 


e210 SPECIES AND QUADRATURE OF 


Therefore, CA=CM. Hence C is the centre of the hyperbola, 
and CO, CQ, are the asymptotes. (Conic Sections, Hyperbo- 
_ la, Proposition XII.) 


Scholium 1. Let EHF represent an eee Say 
hyperbola, and AC, AD the asympto- 
tes ; let the two branches HE and HE 
be brought into the position He and 
Hf, so that the asymptotes become 
A’eand A’f, or till they become parallel 
to each other, and the curve becomes 
a parabola; the parabola then, may be 
regarded as an hyperbola, whose a- «5——7 7H; FD 
symptotes are parallel, and infinitely 
extended in each direction. If the extremities e, f of the curve 
are brought into the positions 7, 7, so as to incline toward the 
axis AB, so that the curve may again return into itself as its 
axis is extended, it then becomes an ellipse or portion of an ~ 
ellipse. These different figures are the result of the position 
of the plane forming the section through the cone. 


Scholium 2. A section of a polyedroid by a plane oblique to 
its axis may assume a combination of one, two, or three, of 
the varieties of surfaces. Thus, a section through a regular 
vertical quadredroid by a plane, parallel to a plane touching 
one of its sides, making an angle of 45° with the axis, toward 
the vertex, will consist of two parabolas on opposite 
sides of the same base, these will evidently be 
parabolas of equal type or similar parabolas, 
when the section passes through the centre 
of the solid, and as it approaches toward one 
of its sides, the parabolas become dissimilar. 


' than 45° with the axis, the section will consist of the segments 


of two ellipses on opposite sides of the same ordinate, which 
ordinate is the line formed by the intersection of the cutting 
plane, with the plane through the centre of the solid perpendi- 
cular toits axis ; and if the cutting plane passes through the 
centre of the solid the two segments will be of similar type, 
but they will vary as the section recedes from the centre 
toward either side. 

If the plane should be passed through so as to make an 
angle with the axis less than 45°, then we should have hyper- 
bolas on the same base, these would be similar hyperbolas 
when the plane should pass through the centre, but would be- 
come dissimilar as it recedes from the centre. 


SECTIONS OF ELEMENTARY SOLIDS. 11 


In like manner, ifa plane should be passed 
through the pentadroid AB parallel to one of 
its sides AD, the section would consist of two 
species, viz., that part which passes through 
the portion ABD would be a parabola, that 
passing through the part ABEC would be a 
segment of anellipse. But if the section should 
be passed so near to BE as to cut the base of the solid, the 
section would consist of a parabola, and a middle segment of 
an ellipse, and would have a rectilinear base. If the plane 
should be parallel to the side EB, then the section through the 
lower part ofthe solid would be a parabola, or a segment of a 
parabola, and the section through the upper part an hyper- 
bola, and in fine, if the plane should make an angle with the 
axis less than that of the side EB, then the section through 
both parts of the solid would consist of the dissimilar hyperbo- 
_ las or segments of hyperbolas. 

Leta plane be passed through a polyedroid of a greater 
number of sides oblique to the axis, and the section may be so 
made as to consist of segments of all the varieties of the conic 
sections, and may also, under certain conditions, have one, 
two, three, and at most four rectilinear sides, but it can have 
no more than two rectilinear sides, except where the section 
is parallel to the axis. 


PROPOSITION IV. LEMMA. 


From a rectangular prism there may be taken two pyramids of 
equal base and altitude with the prism; when there will re- 
main two other pyramids, each equal to half one of the lateral 
sides as a base, multiplied by one third of the distance of such 
base to the opposite side. 


For let the rectangular prism AH be fbi 
divided into twoparts, by passing the ,G77//77 
planes EGDB through the opposite edges, [Sx 
and the portion ABDCGE will consist 
of the pyramid, whose base is ABDC, and 
vertex E, plus the triangular pyramid, 
whose base may be taken as GEC and ver- 
tex D, or DCG may be regarded as the | 7/7 7, 
base, and E the vertex. And since the a©“~ ———B 
other portion of the prism is similar to this, it can be divided in 
a similar manner. Hence the whole prism consists of two 
equal pyramids erected on the upper and lower bases of the 
prism + two other pyramids erected on the lateral sides as 


12 SPECIES AND QUADRATURE OF 


bases, and whose vertices will be in an angle tormed by the 
intersection of its opposite side with one of the bases. Hence 
as in the proposition. 


Cor. 1. If a prism have a square base and a section be 
made through the prism parallel to the base, the sections through 
the pyramids erected on the upper and lower bases will be 
squares, and the sections through the pyramids, whose bases 
are the lateral faces of the prism, will be rectangles whose 
factors are the sides of the squares composing the sections 
through the former pyramids. 

That the sections through the pyramids, erected on the 
square bases will be squares, is sufficiently manifest; and 
since each of the other pyramids coincides with one of these 
along one of its slant sides, and with the other along another 
of its sides, it follows that the measure of its section through 
any parallel portion of the solid will be the rectangle of the 
edges of the section formed by the same plane through the two 
former pyramids. And since, if the section is taken in the 
middle, equidistant between the two bases, the sides of the 
sections through the pyramids with square bases are=half the 
sides of their bases, the sections at such place will each be=1 the 
section of the whole prism ; the sections of the two quadran- 
gular pyramids will be 2 squares=2 quarters of the whole 
section ; hence the sections through the two triangular pyra- 
mids are squares=to the former, and equal to each other, since 
the four pyramids fill the space, and constitute the whole 
prism. 


Cor. 2. Hence, also any section of a prism with a square 
base, made by a plane parallel to such base may be expressed 
by the square of a binomial, whose terms are composed of the 
sides of the sections through the quadrangular pyramids. 
Let hi or hs, the side of the square forming a section 
through the pyramid erected on the lower base, be represented 
by a; and let pd or dn, the side of the square forming a sec- 
tion through the pyramid erected on the upper base, be re- 
presented by 6; then will a+b=the line ho, the side of the 
whole section hodc, and a’ will represent the section through 
the pyramid ABDCE, 6’ will represent the section through 
the pyramid EF HGD, and ad will represent a section through 
each of the other pyramids, hence 2a will represent the sec- 
tions through both, and a’+2ab+0? will represent the whole 
section in whatever parallel the section is taken, always ob- 
serving that the values of a and 6 vary according to the sides 
of the respective sections, which they represent. 


SECTIONS OF ELEMENTARY SOLIDS.  _—18 


Cor. 3. If there be a series of numbers in arithmetical pro- 
gression, whose first term is 0 and last term z, and if the num- 
ber of terms is infinite between these extremes, then the sum of 
the squares of the series of numbers will be equal to one 
third of the square of the last term drawn into the series. 
For if an infinite number of planes be passed through the py- 
ramid, parallel to its base, and equidistant from each other 
through the sides of the sections made by those planes, they 
will be a series of numbers in arithmetical progression, whose 
first term is 0, and the last term may be called z. Now the 
sum of the sections drawn into their distance will represent 
the solidity of the pyramid, but the solidity of the pyramid is 
equal to one third of an equal ‘series of z, the last term or 
base of the pyramid drawn intothe distance or ratio ; or= 
to one third z drawn into the series. 


¥Cor. 4. If there be a series.of numbers in arithmetical pro- 
gression increasing from 0 up to z, drawn into a similar series, 
decreasing from z down to 0, then will the sum of their pro- 
ducts be equal to half the sum of the squares of one of the 
series. For the sections through the pyramid HDGH, as we 
have seen, represent the rectangles of the sides of the corre- 
sponding sections through the two pyramids, formed on the 
two bases, and the sides of these sections are evidently, in each 
pyramid, a series in arithmetical progression ; one increasing 
from 0 to z, while the other decreases from zto0. Moreover, 
the sections.through the pyramid CDGE represent the solidity 
of that body, as the corresponding sections through the pyra- 
mid ABDCH, represents the solidity of that body. But the 
solidity of the pyramid CDGE, is equal to half the pyramid 
ABDCE, which as we have shown, may be represented by 
the sum of the squares of a series of arithmeticals, &c. 


Cor. 5. If in the expression a’+2ab+b’, a be made to pass 
successively through all the values from 0 up to z ; and bat 
the same time pass through all the changes from z to 0; then 
the sum of all the a’ will be equal to the sum of all the 2a), = 
the sum of all the 2’. 


Cor. 6. If a in the expression above, be made to increase 
from / successively to z, and at the same time } be made to 
pass through all the values from z to h, then the series repre- 
sented by 2ab, will be a mean proportional between those re- 
presented by a* and 0’, since in this condition the two pyra- 
mids represented by the series of a? and 8’, would be such as 
pertain to pyramids inscribed in a frustum of a pyramid, and 
erected on the two bases, and the portions represented by the 


14 SPECIES AND QUADRATURE OF 


series of 2ab, would be such as pertain to the pyramids erected 
on the lateral sides as bases, and since the sum of these pyra- 
mids is equal to either of the others, when the two bases 
are equal, and because ab is a mean proportional between a’ 
and 5°, it follows that the series represented by 2ad is a mean 
proportional between those represented by a’ and 6°, which 
agrees with the property of the frustum of a pyramid found 
in the Elements of Geometry. 


Cor. 7. Let any plane KIHL be G N 
passed through a prism parallel to its 
base, cutting the pyramids ABDCEH, © 
EFNGD, DCGE, and EF BD in the sec- 
tions Hnot Krop, Ipot, and Lron. Then # 
since It is equal to Ln or ro, and Hx=on, 


it follows that as the section Hnot : Ipot: : Li »? 
Ipot : Krop : : Hnot : Lnor. Naa 4 


And by addition: : (Knot+Ipot)=HIpn : KLnp. 
Henee Huot: HIpn: : HIpn : KIHL. 


Cor. 8. Hence, of two quantities, the square of the first is to 
the rectangle of the first and second, as the rectangle of the 
efirst and second to the square of the second, and as the sum 
of the first and second x by the first, is to the sum of the first and 
second X by the second. Also as the square of the first is to 
to the sum of the first and second, X by the first, so is the sum 
of the first and second X by the first, to the first and second X 
by the first and second. Thus let aand 6 be two given quan- 
tities, then will a? : ab:: ab: b’, 3: a’+ab: ab+b’, 

and a’: a’?+ab::a+ab: a’+2ab+4b’. 


Scholium. It has been shown (Prop. XXXIV. B II. Hi. Sol. 
Geom.) that the solidity. of a prismoid is equal to the product of 
the sum of the areas of the two ends+4 times AD and EH a mid- 
dlesection, equidistant between them Xj of the altitude; we may 
easily infer that this is also true of all prisms and pyramids 
and pyramidal frusta. Then since the prism AH is equal to 
the sum of the areas of the two bases+4.times the middle 
section hodc x } of the altitude AE, it follows that whatever 
parts make up this product are equal to the whole prism. 


First, then let us take the pyramid ABDCE = (the base 
ABDC-+4 higs, a middle section) x} AE, andalso the pyramid 


SECTIONS OF ELEMENTARY SOLIDS. 15 


EFHGD=(the base EFHG+4gpdn) x 1 s 
AD; if from the expression for the whole E Wy g 
prism we take the expressions for the two [Rw 
pyramids erected on the bases we have, =e 
4sencXiAE+4iopgX}Ak, that is the two , )e==")>.— 
remaining portions, are equal to four times === 
their middle sections, multiplied by } of 
their altitude. Hence we may inferthatif | / 
any portions of either of the pyramidsinto 4™ 
which the prism has been conceived to be 
divided, be cut off by a plane, or planes parallel to the base, 
the portions so cut off will be equal to the product of the sum 
of their two bases, + four times a middle section between 
them, X by 2 of the altitude of such portions. It may be ob- 
served that since a regular pyramid has but one base, its soli- 
dity is hence equal to the sum of this base + four times a sec- 
tion, midway between the base and vertice x } the altitude; and 
also in the two pyramids, whose bases are on the lateral sides 
of the prism, since they have no bases parallel to the middle 
section, their solidities for that reason are equal to four times 
their middle sections x 1 of their height. It may be further 
observed that each portion sgne, GE into which these pyra- 
mids are divided by the plane hodc is a wedge, whose base is 
the section forming the division. 


PROPOSITION V. THEOREM. 


If from the extremity of an ordinate to the axis of a parabola 
and perpendicular thereto, a line be drawn meeting another 
line, drawn from the vertex perpendicular to the axis, form- 
ing with the ordinate and abscissa a rectangle ABCD, and 
if a diagonal be drawn from the vertex A to the extremity of 
the ordinate, forming a right angled triangle ABC of the 
same base BC and altitude AB, then any line or ordinate 
drawn from the axis across the triangle, the parabolic area, 
and the rectangle, parallel to the ordinate, will be cut in con- 
tinued proportion by the sides of those figures. 


That is, EF: EG::EG: EH. 

Or EF, EG, EH are in continued 
proportioh. 

For by (Prop. VII. Cor. of Parabola) g¢ 


hence EF : BC: : EG’: BC’. : 
Or, EF : EH :: EG’: EH’, 
therefore (Prop. XXIV. B. I. El. Geom.) EF, EG, EH, 
are proportionals, or EF : EG: : EG : EH. 


16 SPECIES AND QUADRATURE OF 


Cor. Let any number of ordinates GI be drawn across the 
exterior parabolic space parallel to the axis, and_ since 
EG? : BC?:: AE: AB, by equality we have Al’: AD*®: : IG: DC, 
and since this is true from whatever position on the line AD, 
the line IG may be drawn, it follows that any line IG drawn 
across the exterior parabolic space is proportional to the square 
of the distance of such line from the vertice A. 


PROPOSITION VI. THEOREM. 


If a parallelogram be circumscribed about a parabola, the area 
of the space exterior to the parabola will be equal to one 
third of the parallelogram, and the interior space will be two 
thirds of the parallelogram. 


Let ABCD be aparallelogram cir- 
cumscribed about the parabola 
AEB, and the space ECB, EDA ex- 
terior to the curve will be = } the 
parallelogram ABCD, and the space ek 
AEBA within the curve will be = 2 
ABCD. 


A F B 

From the vertice E draw EF the axis of the parabola, 
which will divide the parabola and rectangle into two equal 
parts; it is to be proved that the exterior space ECB is = 
1 EFBC. 

Draw the diagonal EB, and let an indefinite number of 
equidistant ordinates Ks’ be drawn across the triangle EBC, 
and exterior parabolic space EIBC, and those ordinates 
will represent their respective surfaces in the relation of their 
magnitudes respectively. Then since the distances of those 
ordinates on the line BC, estimated from the vertice B, are a 
series of numbers in arithmetical progression, the ordinates 
Kk drawn across the triangle, are also a series in arithmeti- 
cal progression, for Kk is proportional to BK in whatever 
position the ordinate Kk is drawn; and since it has been 
shown (Prop. V. Cor.,) that any ordinate IK drawn across the 
exterior parabolic space is proportional to the square of its 
distance BK. from the vertex B, it follows that the ordinate is 
proportional also to the square of Kk, the corresponding 
ordinate drawn across the triangle. But the ordinate EC or 
base of the triangle is equal and identical with the base of the 
parabolic exterior space ; hence each of the ordinates IK ter- 
minated by the parabolic curve is equal to the square of K&, 
the corresponding ordinate drawn across the triangle. 


SECTIONS OF ELEMENTARY SOLIDS. 17 


Hence all the ordinates Kk may be represented by a 
series of numbers in arithmetical progression, whose first term 
beginning at B, is infinitely small or 0, and last term EC, and 
all the IK. will be a similar series of squares of those arithme- 
ticals ; but (Prop. IV. Cor. 3,) the sum of an infinite series of 
the squares of a series of numbers in arithmetical progression, 
whose first term is 0, and last term z, or EC, is equal to 4 


EC XxCB, which is 1 of the rectangle EFBC. 


Cor. 1. Hence the parabolic segment EIBE, cut off by the 
diagonal, is equal to 3 of the rectangle, = } of the interior pa- 


rabolic space EIBF, = 1 the exterior parabolic space EIBC. 


Scholium. It has been shown in the argument above that 
the ordinates drawn across the exterior parabolic space are 
severally = the squares of the same ordinates drawn across 
the triangle EBC, this is true where EC=1, and for all other 
values of EC, they are in the same proportion. 


PROPOSITION VII. THEOREM. 


The area of the exterior parabolic space included in its cir- 
cumscribing rectangle is equal to the sum of tts base + four 
times the line or ordinate drawn parallel to the base, and equi- 
distant from the base to the vertex, multiplied by 1 of the al- 
titude. 

And the area of the interior space of a parabola is equal to the 
sum of its base, + four times the ordinate equidistant from 
this base to the vertex, multiplied by one sixth of the altitude. 


Let ABDC be a rectangle circumscribing 
the semi-parabola AcDB, and let Le be an 
ordinate drawn cross the exterior space 
ACDcA equidistant from A to C, and let M 
be an ordinate drawn across the interior 
space AcDB ina similar manner, then will 
the area ACDcA = (CD -P4tie) 1 AC. And 
the area AcDB = (AB+4cM) 3 BD. For 
draw the diagonal AD, then, since we have A 
shown in the argument to Prop. V1, that Le=Lh’, when cD= 
CD’, and since LA=1 CD: then if we call Lh, a, and he 2a, 
we shall have Le = a*, and CD = 4a’ that is Le = 3 CD or 
CD=4Le. 

Hence CD+4Lc=2CD, then by the proposition 20D x }AC 
= CDx 4 AC = the surface AcCDA, as was also found in 
proposition VI. 


18 SPECIES AND QUADRATURE OF 


Again since LM =CD or AB, and Le =1 CD, cM= 3 AB. 
Then by the proposition (4cCM+AB) X } BD=4ABX}{BD= 
2 (ABxBD)=the area, as found in the preceding proposition. 

Cor. 1. Hence the parabolic segment ADcA is equal to 
four times the middle ordinate ch X } the altitude AB. For 
since Lh = 1 LM, ch=Lc = } AB, and } ABX4=AB, and 
ABxX1 BD = the surface which agrees with Cor. 1 Prop. VI. 

Cor. 2. Also if another similar parabolic curve DeA be de- 
scribed on the opposite side of the diagonal, the space between 
the two curves would in like manner be found, = 4 times the 
ordinate ce X 1 of the altitude BD. 


Cor. 3. The same may be shown in reference to the area of 
the triangles ACD, and also of the rectangle ABDC, viz., their 
areas are equal to the sum of their upper and lower bases, + 
4 times a middle ordinate, X } of their respective altitudes. 


Scholium. Let there be a rectangle CB, and a prism BK of 
the same altitude, and let the base of the prism be a square, one 
of whose sides is = AB the base of the rectangle. 

Let the prism be divided Hi 
into the pyramids BEGFD, 
DIKHG, DHFG, EGID, as 
in Prop. IV. Also let the 
parabolic curves. AcD AeD 
and the diagonal AD be 
drawn as in the proposition 
above. Then if a plane 
MzOP be passed through 
the prism parallel to its “ 
base, and if an ordinate LM 
be made to pass through the rectangle in the same plane, the 
plane will cut the pyramidal portions of the prism in the same 
relation, as the ordinate cuts the parabolic portions of the rect- 
angle ; viz., the section Mfsqg through the pyramid BEGCD 
will be to the section MnOP through the prism, as the ordinate 
eM through the exterior parabolic space ABDeA, to the ordi- 
nate LM through the rectangle ABDC ; and the section svO¢ 
through the pyramid DIKHG is to the whole section through 
the prism, as the ordinate Le through the exterior space 
ACDcA, to the whole ordinate LM through the rectangle, and 
so for the sections through the other pyramids, and ordinates 
through the interior parabolic segments, and this is true in 
whatever parallel position the plane forming the section and 
ordinate is drawn. 


For it has been shown (Prop. VI. Sch.) that the ordinate 
eM is equal to hM? ; the section Mfsq is evidently=the square 
of Mf; but Mf is= AM, hence Mfsqg =hM? ; therefore Mfsq 


SECTIONS OF ELEMENTARY SOLIDS. 19 


will be expressed by the same numbers as eM, and this is true 
in whatever parallel position the plane is passed. | 
The same may also be shown in reference to the sections 
svOt through the pyramid DIHG, and the ordinates Le and 
LA, viz., the section svOt=O??=fn? =LA’ and Le=LA*. And 
since every ordinate LM through the rectangle, has the same 
relation to the rectangle, as every section through the prism, 
has to the prism, and each section Mfsg, svOt, has the same 
relation to the solid as each ordinate Le, cM, has to the surface ; 
it follows that the remaining sections, fnvs, stPg, have the 
same relations to their respective figures, and since the sec- 
tions Mfsq svOé are expressed by the same powers of the 
same factors as the ordinates Le and eM are expressed, ch and 
he must also be expressed by the same terms as the sections 
fnvs, stPq, viz., he=LAXhM, ch also=LAXhM, and ce= 
2 (LAXhAM.) Hence if LA =a and hM=6@ then will Le=a’, 
ce=2ab and cM=0", and the whole ordinate may also be ex- 
pressed by a?+2ab+0’, which is the square of a binomial, the 
same as has been shown, (Prop. IV. Cor. 2) in reference to 
the pyramids composing a prism, and the expression is true 
in whatever parallel position the ordinate is drawn. It fol- 
lows therefore that the ordinates drawn across any of the pa- 
rabolic portions of the rectangle, have the same power of de- 
termining the area of the exterior or interior parabolic spaces, 
as the corresponding sections through the pyramidal portions 
of the prism have, in determining the solidities of those pyra- 
mids. But we have shown (Prop. IV. Sch.) that each 
portion of the prism divided as above, however selected or 
compounded, is equal to the sum of its two bases+4 times a 
middle section X ? of the altitude ; hence also the area of any 
parabolic portion or portions of the rectangle is equal to the 
sum of its bases + 4 times a middle ordinate X 3 the altitude. 


Cor. 4. The parabolic area included between two parallel 
ordinates is=the product of the sum of the two ordinates + 4 
times an ordinate equidistant between them X 1 of the altitude 
of the parabolic segment. For let ABMec bea segment of 
the parabola included between the ordinates cM, AB, the axis 
MB and the curve Ac ; draw cS perpendicular to AB, and the 
parabolic segment ScA will be equal to (AS + 4 um) xi Se, 
and the rectangle SBMc is evidently=(S8+cM+4mp) 3 cS ; 
hence the whole space ABMc=(AS+SB+cM+4um-+4mp,) 
xiSce=(AB+cM+4up) 3 Se. 


Cor. 5. In like manner it may be shown that the space 
eMBA is=(AB+eM+4wp) 4MB, and hence also that any para- 
bolic portion AuchmA or uchm may be determined in the same 
manner. 


20 SPECIES AND QUADRATURE OF 


Cor. 6. Hence generally, if a space be terminated by a 
parabolic curve on ene side, and either a parabolic curve or 
aright line on the other, the space included between two 
parallel lines, drawn across the figure cutting those sides, will 
be=to the product of the sum of those lines+4 times another 
line drawn across equidistant between the two multiplied by} 
the perpendicular distance of the two parallel lines. 


PROPOSITION VIII. THEOREM. 


If a circle be cut by a plane through its axis, and perpendicu- 
lars be drawn from every point in the circumference to the 
plane, the orthographic projection of the circle so drawn will 
be an ellipse. 


Let the circle ARML be inclined to the plane of this paper 
in such a manner, that the semicircle ARM may be above 
the paper, and the semicircle ALM below it, and let AM be 
the common intersection 
of the two planes. Let 9 z 
the semicircle ARM be 
projected downwards up- 
on the plane of the paper, 
by drawing perpendiculars 
QP, RB, from each point 
of the eircle, and let the » 
semicircle ALM be 
projected upwards, by 
drawing the perpendicu- 
lars gp, LO, &c. ; then the 
curve ABMO, marked out 
by this projection, will be as 
an ellipse. For draw QN, RC, at right angles to AM, and 
join PN, BC; then the angles QNP, RCB, will measure the 
inclination of the planes, and PN, BC will be perpendicular to 
their common intersection AM. Now QN:: PN: rad. : cos. 
QNP, and RC : BC: rad. : cos. (RCB = QNP) ; 
“. QN:PN:: RC or AC: BC. which agree with the pro- 
perties of the ellipse (Prop. XII. Cor. Conic Sec.) Ina similar 
manner it may be shown that the semicircle ALM is pro- 
jected into a semi-ellipse AOM ; and thus the whole circle 
ae is projected into an ellipse ABMO, whose major axis 
is 


Scholium. This proposition is manifestly true, when the 
plane of the projection does not cut the circle, or cuts it un- 
equally. 


SECTIONS OF ELEMENTARY SOLIDS. 21 


Cor. Hence any section of a cylinder by a plane not per- 
pendicular or parallel to its axis, is an ellipse, or portion of an 
ellipse. | 


PROPOSITION IX. THEOREM. 


If on the major axis of an ellipse a circle be described, the area 
of the ellipse will be to that of the circle, as the minor axis of 
the ellipse to the major axis. ; 


Let ACBD be a circle described on the major axis AB of 
the ellipse AEBF, then will the area of the ellipse be to that 
of the circle as EF to EB. 

For it has been shown 
(Prop. XII. Cor. Conic 
Sec.) that any ordinate IL 
of the axis AB of the ellipse 
is to the corresponding 
ordinate GH of the circle, 
as EF to CD or AB, as the 
minor axis to the major 
axis. And since this is 
true in whatever position 
the ordinates to this axis are 
drawn; and because if we 
suppose an indefinite num- 
ber of equidistant ordinates 
to be drawn across the ellipse and circle, the sum of the ordi- 
nates in each will represent those figures in the relation of their 
areas, the sum of the ordinates in the ellipse will be to the sum 
of those in the circle, as EF to CD, and hence the area of the 
ellipse will be to that of the circle in the same ratio. 


Scholium. As the area of the ellipse bears this given ratio to 
that of its circumscribing circle, the quadrature of the ellipse 
must therefore depend on the quadrature of the circle. If 
*xX(AQ. AO, or CO,) or rAO*=the area of the circle, then 
«.AO.KO=the area of the ellipse. Hence the area of the 
ellipse is found by multiplying the rectangle under its semi- 
axis by the constant number 7, the ratio of the diameter to 
the circumference of a circle, which in the Elements of Geo- 
try we have found, developed to a certain order of decimals 
to be 3,1415926. But 3,1416 may be regarded as the 
value of « which is a convenient number to use, and is suffi- 
cient where extreme accuracy is not required. 

From this it also appears, that the area of an ellipse is 
equal to the area of a circle whose radius is a mean propor- 


22 SPECIES AND QUADRATURE OF 


tional between its semi-axis; for the area of that circle is = 
x R’=7 Xthe square of /AOxEO = 7. AO.EO. 


Cor. 1. The area of an ellipse has the same ratio to the 
area of its circumscribed parallogram as the area of a circle 
has to its circumscribed square. For the area of the parallelo- 
gram circumscribing the ellipseis = 4 AOX EO; hence the 
area of the ellipse: area of the parallelogram: : +. AQ. EO: 
4AO.EO:: 83,1416: 4::'7854: 1. 


Cor. 2. If a circle be described on the minor axis EF of the 
ellipse, then any ordinate PQ of the circle will be to the cor- 
responding ordinate Ie of the ellipse as EF to AB. Hence 
also the area of the ellipse is to that of the circle described on 
its minor axis, as the major, axis of the ellipse to its minor 
axis. 


Cor. 3. Hence any segment of the ellipse cut off by ordi- 
nates, either of the major or minor {axis is to a similar seg- 
ment of the circle, described on such axis, as its conjugate is 
to such axis. Thus the segment IAL of the ellipse AEBF is 
to the segment GAH of the circle described on the axis AB, 
as the ordinate IL is to the ordinate GH or as the minor axts 
EF of the ellipse is to the axis CD of the circle, or to AB the 
transverse axis of the ellipse. And the segment Ie is to the 
segment PQE as the axis AB as conjugate to EF of the ellipse, 
to the axis RS of the inscribed circle, or EF of the ellipse. 


Cor. 4. Since the area of a circle BESF is equal to its cir- 
cumference multiplied by half the radius EO, and since the 
area of the ellipse AE BF is = to that product increased in the 
ratio of EO to AO or CO, it follows that the area of the 
ellipse is equal to the circumference of its inscribed circle mul- 
tiplied by 3 the radius of its circumscribed circle, and that it 
is also equal to the circumference of the circumscribed circle 
multiplied by 4 the radius of the inscribed circle. 


Cor. 5. As a circle is to the square of its diameter so is 
any ellipse to the rectangle of its two axes, or the rectangle 
of any two conjugate diameters drawn into the sine of their 
included angle. Any two like segments or zones of the circle 
and ellipse, are alsoin the same proportion. 


SECTIONS OF ELEMENTARY SOLIDS. 23 


Cor. 6. Hence we have the fol- 
lowing construction ; let ADE be an 
oblique segment of the ellipse 
AFBGA, cut off by an_ ordinate 
to the diameter AB, FG being 
_ the conjugate. Through the 
centre C draw aP perpendicular 
to FG meeting Aa and BP, in a and 
P, both parallel to FG ; then about 
the axes a P, FG, describe the ellipse 
aFPGa, meeting the ordinate pro- 
duced in eand d. Thenwill the right 
elliptical segment dae, be equal to the oblique segment DAE, 
and so will the whole ellipse ak PGa=AFBGA ; moreover 
their corresponding ordinates de DI: parallel to the common 
diameter FG, are everywhere equal, as are the like parts or 
zones contained between any two of such ordinates. And the 
same may be said of all ellipses of the same diameter FG, con- 
tained between the parallels aA, BP infinitely produced. 


. PROPOSITION X. THEOREM. 


All the parallelograms are equal, which are formed between the 
asymptotes and curve, by lines drawn parallel to the a symp- 
totes of an hyperbola. 


That is, the lines GE, EK, AP, AQ, 
being parallel to the asymptotes CH, Ch, 
then the parallelogram CGEK = paral- 
lelogram CPAQ. 

For, let A be the vertex of the curve, 
or extremity of the semi-transverse axis 
AC,: perpendicular to which draw AL 
or Al, which will be equal to the semi- 
conjugate, by definition XIX. Hyp. Also, 


draw HEDedA parallel to Iv. ahs bag. 
Then, CA?: AL? 3: CD CA? : DE’, 
and by parallels, At Abang Gers 


therefore, by substract. CA? : AL’ :: CA’* : DH?—DE? , 
or rectangle HE . HA ; 
consequently, the square AL’=the rectangle HE. EA. 
But, by similartrian. PA: AL:: GE: EH, 
and, by the same, QA: Al:: EK: Eh; 
therefore, by comp. PA. AQ: AL’?:: GE. EK : HE. Eh; 
and, because AL’=HE . EA, therefore PA. AQ=GE. EK. 
But the parallelograms CGEK, CPAQ, being equiangular, 
are as the rectangles GE.EK and PA . AQ. 


24 SPECIES AND QUADRATURE OF 


And therefore the parallelogram GK = the parallelo- 
gram PQ. 

That is, all the inscribed parallelograms are equal to one 
another. 


Cor. 1. Because the rectangle GEK or CGE is constant, 
therefore GE is reciprocally as CG, or CG: CP: : PA: GE. 
And hence the asymptote continually approaches towards the 
curve, but never meets it ; for GE decreases continually as 
CG increases ; and it is always of some magnitude, except 
when CG is supposed to be infinitely great, for then GE is in- 
finitely small or nothing. So that the asymptote CG may be 
considered as a tangent to the curve at a point infinitely dis- 
tant from C. 


Cor. 2. If the abscissas CD, CE, CG, 
&c., taken on the one asymptote, be in 
geometrical progression increasing ; then 
shall the ordinates DH, EJ, GK, &c., pa- 
rallel to the other asymptote, be a de- 
creasing geometrical progression, having 
the same ratio. For, all the rectangles if 
CDH, CEI,CGK, &c., being equal, the 
ordinates DH, EI, GK, &c., are reciprocally as the abscissag 
CD, CE, CG, &c., which are geometricals. And the recipro- 
cals of geometricals are also geometricals, and in the same 
ratio, but decreasing, or in converse order. 


Be 


PROPOSITION XI. THEOREM. 


The three following spaces, between the asymptotes and the curve, 
are equal ; namely, the sector or trilinear space contained 
by an arc of the curve and two radit, or lines drawn from 

- its extremities to the centre ; and each of the two quadrila- 
terals, contained by the said arc, and two lines drawn from 
ats extremities parallel to one asymptote, and the intercepted 
part of the other asymptote. 


That is, 

The sector CAE =PAEG=BAEK, 
all standing on the same arc AE 

For, as has_ been already shown, 
CPAB=CGEK ; 

Subtract the common space CGIB, 
ee shallthe parallel PI=the parallel 
. Toeach add the trilineal IAE, 


SECTIONS OF ELEMENTARY SOLIDS. 25 


Then is the quadril. PAEG=BAEK. 

Again, from the quadrilateral CAEK, take the equal triangle 
CAB, CEK, and there remains the sector CAE=BAEK. 

Therefore, CAE=BAEK=PAEG., 


PROPOSITION XII. THEOREM. 


Every inscribed triangle, formed by any tangent and the two 
intercepted parts of the asymptotes, is equal to a constant 
quantity ; namely double the inscribed parallelogram. 


That is, the triangle CTS=2 parallelogram GK. 

For, since the tangent TS is bisected 
by the point of contact E, and EK is pa- 
rallel to TC, and GE to CK ; therefore ye - 
CK, KS, GE are all equal, as are also 
CG,GT, KE. Consequently the triangle ¢ x = 
GTE = the triangle KES, and each 
equal to half the constant inscribed paralielogram CK. And 
therefore the whole triangle CTS, which is composed of the 
two smaller triangles and the parallelogram, is equal to 
double the constant inscribed parallogram GK. 


PROPOSITION XIII. THEOREM. 


If from the point of contact of any tangent, and the two inter- 
sections of the curve with a line parallel to the tangent, three 
parallel lines be drawn in any direction, and terminated by 
either asymptote ; those three lines shall be in continued 
proportion. 


That is, if HKM and the 
tangent IL be parallel, then are 
the parallels DH, EI, GK in x 
continued proportion. 


Pe Be’ 


C aks Be Got, M 


For, by the parallels, EI: IL ::DH: HM; 
and, the same EI: IL::GK:KM; 
therefore by compos. EI? : IL?:: DH. GK: HMK; 
but, the rect. HMK=IL? ; 
and therefore the rectangle DH . GK=EI’, 
or DH: El :: EI: GK. 
3 


&: 


» 


26 SPECIES AND QUADRATURE OF 


PROPOSITION XIV. THEOREM, 


Draw the semi-diameters CH, CIN, CK ; 
(see last diagram.) 
Then shall the sector CHI = the sector CIK, 


For, because HK and all its parallels are bisected by CIN, 
therefore the triangle CNH=trian. CNK, 
and the segment INH=seg. INK ; 
consequently the sector, ClH=sec. CIK. 


Cor. If the geometricals DH, El, GK be parallel to the 
other asymptote, the spaces DHIE, EIKG will be equal ; for 
they are equal to the equal sectors CHI, CIK. 

So that by taking any geometricals CD, CE, CG, &c., and 
drawing DH, EI, GK, &c., parallel to the other asymptote, as 
also the radii CH, CI, CK ; 

then the sectors CHI, CIK, &c. 

or the spaces DHIE, EIKG, &c. 
will be all equal among themselves. 
Or the sectors CHI, CHK, &c. 

or the spaces DHIE, DHK, &c. 
will be in arithmetical progression. 

And therefore these sectors, or spaces, will be analogous 
to the logarithms of the lines or bases CD, CE, CG, &ec. ; 
namely CHI or DHIE the log. of the ratio of CD to CE, or 
of CE to CG, &ec. ; or of El to DH, or of CK to EI, &c. ; 
and CHK or DHKG the log. of the ratio of CD to CG, &c. 
or of CK to DH, &c. 


Scholium. If the common logarithms are multiplied by 
2,302585093 their products will be the hyperbolic logarithms. 


PROPOSITION XV. THEOREM. 


If any number of ordinates AD, EF, &c. to GH, between the hy- 
perbola and the asymptote, are taken in geometrical progres- 
sion increasing, then will their distances AE, &c., on the 
asymptote be a series of geometricals decreasing, and AG 
will represent the sum of the decreasing series. 


“d 


» > 


SECTIONS OF ELEMENTARY SOLIDS. 27 


That the geometrical a_? 
ordinates and their dis- 
tances on the asymptote 
are reciprocal geometri- 
cals is manifest from(Prop. 
XIV Cor.,) and since the 
distances AE, EK &c., 
on the asymptote, inter- 
cepted by those ordinates 
compose the whole series 
of decreasing geometri- 
cals, thesum of that series 
will be represented by 
AG, the part of the asymptote taken up by the series. 


Cor. 1. The last term Gi, is=AE X EF+GH. For the area 
ADFE = the area HGit, and if the distances on the asymp- 
tote are taken indefinitely small, ti may be regarded as =GH, 
and AD may be regarded as = EF, and if the hyperbola is 
equilateral, then the space ADFE will bb=AEX AD=GixGH 
converting this equation into a proportion, 

we have AK: Gi:: GH: AD, 
or GH: AD:: AE: Gi. 
Hence, Gi=AEX(AD or EF)+GH 


Scholium. It is evident that this proposition is true, however 
far the series may extend on the asymptote CA. 


Cor. 2. If the series commence at any point A on the asymp- 
totes, and decrease to infinity, CA will be the sum of the 
series ; hence the sum of an infinite series of decreasing geo- 
metricals is a finite number. 


PROPOSITION XVI. THEOREM. 


If any number of portions AK, EK, &c., to G, (sce diagram 
above) are the reciprocals of a series of ordinates AD, EF, 
KL, &c., to GH, in geometrical progression ; then will the 
area or space intercepted by AD and GH, between the curve 
and asymptote be = the space ADFE, multiplied by the 
number of terms of the series. 


For since the space intercepted by each two consecutive 
ordinates of the series is equal to ADFE, it follows that the 
whole space intercepted by all the ordinates, or the space 


ADHG is = ADFE x by the number of terms in the series. 


28 SPECIES AND QUADRATURE OF 


Cor. Hence the area of the interior hyperbolic space HDB 
may be found, if the exterior space is known. For the interior 
space is equal the triangle HBD + triangle HnD + the rect- 
angle nDAG—the exterior hyperbolic space GHDA. 


Scholium. Let AD=), EF = e, GH = g, and AK the first 


term of a geometrical series=a ; the last term = z ; and AG 
the sum = s. Then we shall have a the first term, and z a 


the last term, and s the sum of all the terms of a geometri- 
cal series to find the ratio and number of terms. 
Then by geometrical progression will 

s=a-+- ar-++ar + ar® +. >. tap 
multiply by r 

srartar ape oe he 
Subtracting the first from the second, 

s (r—1)=ar"— a 


tt. bare a hare 


__a(r® — 1) 
tl 
And since, 
z=are— 
we r2z—da 
agecod 
multiplying by r—1,  sr—s=rz—a 
: bigivy s—a 
By transposing and dividing;r= —— 


and in the equation z=ar*—* 


multiplying by 7 and dividing by a we have 


2r 
Lo alg caer which equation 


is irreducible by the ordinary methods when zx is the un- 
known quantity, it being an exponential equation, and can be 
solved only by logarithms, which see. (Chap. V Trigonome- 
try) ; but if we proceed to raise 7 to some power, whose 


value is— the index of that power will be the value of 7 ; this 


can always be done to approximate exactness. 

Let the series of geometricals commence at the vertex H, 
making Gi the first term = a, and Jet n be a given number, 
then assuming any value tos, z, r, each of the others may be 
found by the formula above, and hence the area of the exterior 
and interior hyperbolic spaces may be found, as well as that of 
any section or segment. 


SECTIONS OF ELEMENTARY SOLIDS. 29 


Scholium 2. If the portion of the hyperbolic curve is small, 
the ordinates AD, EF, &c., and consequently AE, EK, &c., 
will be very nearly in arithmetical progression as may readily 
be seen by the rapid manner in which the ordinates DN, DV, 
DH, DU, &c., approximate toward the curve as the hyperbo- 
lic arc is decreased, hence the hyperbolic area may be approx- 
imately obtained by first dividing it into portions having 
known ratios to each other, and finding the value of one when 
the whole will become known. 


PROPOSITION XVII. PROBLEM. 


Let it be required to find how far from G on the asymptote AC 
an ordinate must be drawn parallel to GH in order to inter- 
cept with GH an area =A 3 


Take any distance Gi (see fig. to Prop. XV,) and since AG : 
it: : GH: Gi we have, the two parallel sides GH, zt, and their 
distance Gi of the quadrilateral GH to find its area, which 


call f, then == 2, let a represent the distance, Gi the first term 
of the geometric series, and GH ~+ tz = 7, the ratio. 
7h — ] a 
Whence we have s =a (—— = the distance from G 


on the asymptote, that the ordinate must be drawn to intercept 
with GH, the area A. 


Scholium. Since the spaces intercepted by the ordinate GH 
or AD, with any base IN, are the hyperbolic logarithms of 
the ratios of AD to GH (P. XIII. Cor.) hence the area may be 
found by taking the logarithms of the ratio, and multiplying it 
into the base. 


PROPOSITION XVIII. THEOREM. 


If ail the ordinates to the diameter of an equilateral hyperbola 
be reduced in any ratio r, then the area of the hyperbola 
will be reduced in the same ratio. 


Let all the ordinates HI, NO, &c., of H 
the equilateral hyperbola HFI be re- [\ 
duced to hi, no, &c., forming a hyperbola 
hFi, then will the area of the equilate- 
ral hyperbola HII be to the area of the 
new hyperbola hFi as HI to hi, or as 
NO to zo. For, if an indefinite num- 
ber of equidistant parallel ordinates be 
drawn across the two hyperbolas, the 


30 SPECIES AND QUADRATURE OF 


areas of each will be as the sum of the lengths of all the ordi- 
nates in each, but by hypothesis, each of the ordinates HI, 
NO, are to their corresponding ordinates ni, no in the given 
ratio r; hence, their sums in each, must also be in the same 


ratio. 


Cor. 1. Since the areas of similar figures, are as the squares 
of the lines similarly drawn in each; similar segments of simi- 
lar hyperbolas, are as the squares of their axes, or of their like 
diameters. 


Cor. 2. Hyperbolas of the same transverse axis and abscissa 
are to each other as their conjugate axes; but if their bases, 
or ordinates, and conjugate axes be the same, they will be as 
their transverse axes; and generally, hyperbolas having the 
same abscissa, are as the rectangle of their major and minor 
axes ; and, consequently, having the quadrature of any one 
hyperbola, we may from it find that of any other. Thus 
knowing the area answering to any abscissa in any one hyper- 
bola, we can find a similar abscissa in the other; then as the 
rectangle of the axes of the squared hyperbola, is to the rect- 
angle of the proposed one, so is the area of the former, to that 
of the latter. 


Scholium. 1. It may be shown thata parabola is always greater 
than a triangle of equal base and altitude, and that the hyperbola 
is always between the two; since from the nature of the sec- 
tion through the cone by which the hyperbola is produced, it 
may vary from a vertical section through the cone forming a 
triangle, to the section parallel to one of its sides forming a 
parabola, each of which extremes it can never reach, but may 
approach infinitely near. 

When the section passes through the 
vertice and base of a cone, the section 
is evidently a triangle, whatever angle 
it may make with the axis. Let the sec- 
tion be removed from the vertice by 
any quantity however small, and it becomes 
one of the curves described above; it may 
be an hyperbola or a parabola, either of 
which may agree infinitely near with the GE en B (yon FC 
triangle. Hence, the triangle and parabola are not opposite ex- 
tremes of the hyperbola as the limits of its dimensions, for the 
hyperbola and parabola may both terminate ina triangle at the 
same moment, or the two curves may become assimilated when 
they are infinitely removed from a triangular form ; nevertheless, 


when the hyperbola hh, parabola PP, and the triangle ACC are 


AS A A’ 


SECTIONS OF ELEMENTARY SOLIDS. 31 


described on the same base C C, and of equal altitude B A, the 
hyperbola may always be described between the triangle and 
parabola; this arises from the nature of the sections from 
which they are formed. Let the axis be indefinitely extended 
as when the section becomes parallel to one of the sides of the 
cone, in which case the two asymptotes AC, AC of the hyper- 
bola CHC, become parallel to each other, as A’ e, A’ f, in which 
case the asymptotes may be conceived to vanish, and the 
curve, at this point, assumes other properties, as we have seen 
in our investigations ; the axis, after being increased to infinity 
in the direction, BA, vanishes or becomes changed, and is in- 
finitely extended in the direction B; but the same cause that 
brings the asympotes to a parallel position, may, by continuing, 
cause them to assume a greater distance toward A’ A’ than 
in the opposite direction, and the axis in such case becomes 
a definite magnitude, in which case the curve will be inclined 
inward, as at n 7, when, by extending the axis, it will return 
into itself and become an ellipse. 


Scholium 2. As it has been shown (Prop. IX Cor.6) in re- 
ference to ellipses, it may be also shown that all hyperbolas 
having the same centre and equal bases, and described be- 
tween the same parallels, although infinitely produced, are 
equal to each other, as are also their corresponding sections 
parallel to the bases, and likewise any frustum or segments 
intercepted by the parallels. 

The same may also be shown of parabolas of the same base 
and altitude, or those with the same base, and between the 
same parallels, that they are equivalent in area, &c. 


EQUATIONS TO THE CURVES FORMED BY SECTIONS OF THE CONE. 


1. For the Ellipse. 


Let d denote AC, the semi-axis major or semi-diameter ; 

c = CM its conjugate ; 

« = AK, any abscissa, from the extremity of the diam. 

= DK the correspondent ordinate. 
Then, seers Xl Ellipse,) AC? : CM? :: AK . KB: DK, 
that is, d? : c? ::@ (d—x): y®, hence d? y? =c? (d—2?) or dy 
Uae the equation of the curve. 

And from these equations. any of the four letters or quan- 
tities. d, c, x, y, may easily be found, by the reduction of equa- 
tions, iyi the other three are given. 

Or, if p denote the parameter, = c? + -d by its definition ; 
then, (Prop. XII. Cor. 1)d:p: : 2 (d—2): y’, or dy’=p (dv—xz’) 
which is another form of the equation of ‘ curve. 


32 SPECIES AND QUADRATURE OF 


Otherwise. Or, if d = AC the semi-axis;c= CH the semi- 
conjugate ;p = c? +d the semi-parameter ; x = CK the ab- 
scissa counted from the centre; and y = DK the ordinate as 
before. 

Then is AK = d—2x, and KB = d +2, and AK. KB = (d—2) 
X (d+z) = d?—x?. 

Then, d? :c? :: d?— «2: y?,and d?y? = c?(d’—2?), or dy 
=c./(d’—«"), the equation of the curve. 

Or, d: p::@—2: y’, and dy’=p (d?—2°) another form of 
the equation to the curve ; from which any one of the quanti- 
ties may be found, when the rest are given. 


2. For the Hyperbola. 


Because the general property of the opposite hyperbolas, 
with respect to their abscissz and ordinates is the same as 
that of the ellipse, therefore the process here is the same 
as in the former case for the ellipse ; and the equation to the 
curve must come out the same also, with the exception of the 
signs which are sometimes changed from + to —, or from — 
to +, because the abscissz on the axis, lie beyond or without 
the curve, whereas they lie within it, in the ellipse. Thus mak- 
ing the same notation for the whole diameter, conjugate, ab- 
scissa, and ordinate, as at first in the ellipse ; then, the one 
abscissa AK being z, the other BK will be d + 2, which in the 
ellipse was d—— x; so the signofz must be changed in the 
general property and equation, by which it becomes da’: c’: : 
«© (d-2z): y*?; hence d? y’? = c? (dv+2’) and dy = c/(dz+2’), 
the equation of the curve. 

Or, using p the parameter as before, it is, d:p::x(d+z): 
y’, or dy’=p (dx+z’), another form of the equation of the 
curve. 


Otherwise, by using the same letters d, c, p, for the semi-axis, 
semi-parameter, and parameter, and z for the abscissa CK 
counted from the centre ; then AK=a—d, and BK=2x+d and 
the property d’: c’?: :(¢—d) X (a+d): y’, gives dy? = c” 
(x°—-d?) or dy = c./(x’—d’), where the signs of d? and z? are 
changed from what they were in the ellipse. 


SECTIONS OF ELEMENTARY SOLIDS. 33 


Or, again using the semi-parameter, d : y:x*—d?: y’, and dy 
=p (x?—d’) the equation of the curve. 

But for the conjugate hyperbola, (Prop. XII, Cor. 3, Hyp.) 
as the signs of both 2? and d’ will be positive ; for the pro- 
perty in that case being CA’: CB’: : CD?+CA’: Dp’, itis d’: 
c::a+d?:y=Dp’, or d?y’=c" (x*+d’) and dy=c /(x’+a’), 
the equation to the conjugate hyperbola. 

Or, asd: p::a°+d’: y’, and dy’=p («*+d’), also the equa- 
tion to the same curve. 


On the Equation to the Hyperbola between the Asymptotes. 


Let CE and CB be the two asymptotes to the 
hyperbola dFD, its vertex being F; and EF, bd, | 

AF, BD, ordinates parallel to the asymptotes. Put L 

AF or EF=a, CB=z, and BD=y. Then, (Prop. 

XI Hyp.,) Ab. EF =CB.. BD, or @’=zy, the 

equation to the hyperbola, when the abscisse and 4 | \n 
ordinates are taken parallel] to the asymptotes. Sens 


8. For the Parabola. 


If x denote any absciss beginning at the vertex, and y its 
ordinate, also p the parameter. Then 
AK: KD::KD:p:orz:y:: y: p;hence px=y’ is the equa- 
tion to the parabola. 


4. For the Circle. 


Because the circle is only a species of the ellipse, in which 
the two axes are equal to each other ; therefore making the 
two diameters d and c equal in the foregoing equations to the 
ellipse, they become y’=dz—z’ y being the mean proportional 
between z andd— zx, when the abscissa x begins at the 
vertex of the diameter : and y* = d? — z’*, when the abscissa 
begins at the centre. 


Scholium. In each of these equations, we percieve that they 
rise to the 2d or quadratic degree, or to two dimensions ; 
which is also the number of points in which every one of these 
curves may be cut by aright line. Henceitis that these four 
curves are said to be lines of the 2d order. And these four 
_are all the lines that are of that order, every other curve being 
of some higher, or having some higher equation, or may be 
cut in more points by a right line: 


BOOK Ii. 


SOLID SECTIONS OR SEGMENTS OF SOLIDS OF REVOLUTION, 
CYLINDROIDS, AND PARABOLIC PRISMOIDS AND UNGULAS. 


DEFINITIONS. 


1. Solid Sections or segments are the portions of a solid cut 
off, or out, from another solid by one or more plane or curve 
surfaces. 

2. If any portion ofa cylinder oracone, is cut off by a plane 
which is not parallel to the base, the solid section, so,cut off, is 
called an ungula. 


3. Ungulas cut from a cylinder are 
called cylindric ungulas. 

Thus the section ACDBEF cut off 
by the plane ACDB: as also the se- 
veral sections ACIHGM, GHIK and 
HIKLBD, are cylindric ungulas. | 


4. Ungulas cut from a cone or conic 
frustum are called conical ungulas. 
Thus the sections ECDB and AFEGAB 


are conical ungulas. 


5. Portions cut from a sphere by 
the intersection of two planes, are 
called spherical wedges or ungulas. 

Thus the spherical section 
ACBDAB is a spherical wedge or 
ungula, 


SEGMENTS AND UNGULAS. 35 


6. A portion cut from a segment of a 
sphere by a plane perpendicular to the 
base of the segment is called a second 
segment. 


7. Conical ungulas take particular names according ‘to 
the figure of the superficial section, viz., parabolic elliptical 
or hyperbolic. . 

8. The portion of a cylinder or cone remaining after an un- 
gula is taken, is called the complement of the ungula, or ungu- 
lical complement, and its altitude is equal to that of the un- 

ula. 

Thus GNPKIHG (Fig. at Def. 3) is the complement of the 
ungula GKIH, and ADCEHA (Def. 4) is the complement of 
the ungula ECDB. 

9. An ungulical supplement is what remains of the whole 
solid after an ungula is taken therefrom. 

10. An elliptical cylinder is a cylindrical solid, every super- 
ficial section of which by planes perpendicular to the axis are 
equal ellipses. 

11. An elliptical cone is one, every section of which per- 
pendicular to its axis are similar ellipses, and is the solid in- 
cluded between an elliptical base perpendicular to its axis, and 
a point as its vertex. 

12. A cylindroid is a solid included between two bases of 
equal perimeter, one of which is an ellipse, and the other an 
ellipse of a different excentricity or a circle. 

13. If a solid have two parallel bases, consisting of dissimi- 
lar ellipses of different perimeters, or one elliptical and one 
circular base, of different perimeters. The solid may be 
called a conoidal frustum. : 

14. A parabolic prism is a prismatic solid, whose base is a 
parabola, and each of whose sections parallel to the base is 
equal and similar to the base. 

15. A parabolic pyramid is a pyramidal solid, whose base 
is a parabola; and is the solid included between such base, and 
a point above as its vertex. 


36 SEGMENTS AND UNGULAS. 


PROPOSITION I. ‘THEOREM. 


If a cylinder be cut by a plane parallel to its axis, the solid 
section so cut off will be equal to the area of its base multi- 
plied by its altitude, and its curve surface will be equal to 
the arc of its base multiplied by its altitude. 


‘Let HFCGED, be a section of the cylinder, 
AD, cut off by a plane HFEG, parallel to the a 
axis of the cylinder ; then will the solidity of the 
section be equal to the area GD of the base mul- 
tiplied by the altitude DC, and its convexsurface q 
will be equal to the arc EDG of the base multi- 
plied by the altitude DC. 

For, if a plane be passed through a cylinder ,, 
parallel to the axis, it will divide the two bases 
proportionally, and every section dhec parallel 
to the base will be divided in the same ratio; hence the curve 
surface of the cylinder will be divided in the ratio that the cir- 
cumference of the base is divided. But the convex surface 
is equal to the circumference of its base multiplied by its alti- 
tude, (Prop. I. B. II. Hl. S. Geom.,) and its solidity is equal to 
the area of its base multiplied by its altitude, (Prop. I. B. IIL 
El. S. Geom.) Hence the curve surface of the portion so cut 
off, proportional to the section of the base, is also equal to the 
arc of its base multiplied by its altitude; and its solidity, for 
the same reason, is equal to the area of its base multiplied by 
its altitude. 


PROPOSITION Il. THEOREM. 

If a plane cut a cylinder diagonally, passing through the oppo- 

site edges of the two bases, then the cylinder will be divided 

into two equal ungulas, and the curve surface of each ungula 

will be equal to the perimeter of its base multiplied by half 

its altitude ; and the solidity of each will be equal to the area 
of the base multiplied by half the altitude. 


Let ALDMA be a plane passing diagon- 
ally through the cylinder BC, cuttingthe op- » < 
posite edges of the two bases at A and D, 
and the two ungulas ADB, ADC will be 
equal in surface and solidity, and the curve 
surface of each is equal to the circumference 
of its base multiplied by half its altitude, andy 
their solidities are equal each, to the area of 
its base multiplied by half its altitude. 

For, the two ungulas are symmetrical 
(Def. 19, B. II. HZ S. Geom.,) being so- =< 
lids, similarly formed, in opposite sides of BN. 


the plane ALDMA as a base, and hence 


SEGMENTS AND UNGULAS. 37 


are equal each to each. Moreover, let a plane be passed 
through the cylinder, parallel to,.and at equal distances be- 
tween the two bases, forming the superficial section ILKMI, 
and the two ungulas will be cut by that plane in the same 
ratio, since they are similar solids, and since they are cut by 
a plane parallel to and at equal distances from their bases: 
hence the solid section, or partial ungula ALMI, is equal to 
the partial ungula DMLK, both in surface and solidity; and 
for the same reason, their complemental ungulas ALMKC, 
ILMDB are equal, each to each, both in surface and solidity. 
Therefore, the cylindric section KB between the two parallels 
ILKM and BEDF, is equal to the ungula ABEFDA, both in 
surface and solidity; but the cylindric surface of KB is equal 
to the circumference BEDFB of the base multiplied by IB, 
equal to half the altitude of the ungula ABD; and the solidity 
of KB is equal to the area BEDF multiplied by 1B, equal to 
half the altitude of the ungula. Hence, &c. 


PROPOSITION III. THEOREM. 


[fa cylinder be cut diagonally by a plane which bissects the 
base, the ungula cut off by such plane will be equal to its 
cylindric surface multiplied by one-third of the radius of the 
circle of the base. 


Let AFEC be an ungula, cut off from the , 
cylinder ABDC by the plane AFH, bisecting 
the base CFDE in FE, then will the ungula 
AFEC be equal in solidity to its cylindric 
surface multiplied by one-third of the radius 
IC of the base. 

For, conceive the ungula to be divided in- 
to cylindrical elementary pyramidals (Prop. 
XIIL. Schol. 1 and 4, B. Ill. L7. 8. Geom.) by 
planes parallel to the axis of the cylinder, and p 
all passing through the centre I of the base, F 
- then these will all be perfect pyramidals, since 
lines drawn from every point in the cylindrical base to the 
centre I, lie wholly in the solid, or in the plane surfaces of the 
solid; and since all parts of the solid are included between 
the cylindrical base of the ungula and centre I, in right lines. 
And these pyramidals (Prop. XIV. B. III. Ei. S. Geom.) are 
each equal to its cylindrical base multiplied by one-third of its 
altitude, viz., one-third of CI. Therefore, the whole ungula, 
being made up of all the pyramidals, is equal to the sum of all 
their bases multiplied by one-third of the common altitude CI. 


38 SEGMENTS AND UNGULAS. 


Cor. 1. If a line ec be passed along the axis of the cylin- 
der, and the edge of the ungula, through its whole extent, be- 
ing always perpendicular to the axis, in every position ec or 
gh, or an the solid included within the surface described by 
the motion of this line, together with the ungula, is equal to the 
cylindric surface of the ungula multiplied by half the radius 
of the base. 

For let the cylindric surface be divid- ? 
ed at pleasure, by planes passing through @ 
the axis, and the several divisions will , a 
all be elementary portions of the cylinder, || 
and (Prop. II], Cor. B. Ill, Hl. S. Geom.) 
each will be equal to its cylindrical base x 
multiplied by half its altitude, or half the C B 
radius of curvature of the cylinder. 


dl 


Cor. 2. Hence the section included between the ungula 
and the axis of the cylinder is equal to half the ungula ; since 
the ungula, is equal by the proposition to its cylindrical 
surface, multiplied by one-third of the radius of the base, 
and the two sections together, are equal to the same sur- 
face multiplied by 1+-4=4 the same radius. 


PROPOSITION IV. THEOREM. 


From the complement of two similar ungulas whose bases are 
together equal to the base of the cylinder, a cone may be taken 
of equal base and altitude to that of the cylinder, when there 
will be left a residual portion equal to its cylindric surface 
multiplied by one-third of the radius of the base. 


Let DEFC be the complement of the two 
similar ungulas DEF A, and CEFB whose 
bases AEF and BEF are together equal to 
the base of the cylinder, and there may be 
taken a cone DGCHL of equal base and alti- 
tude with the cylinder, and the portion of the 
cylinder remaining will be equal to its cylin- 
dric surface multiplied by one-third of the 
radius IF of the base 


For since each of the ungulas are cut off by planes pass- 
ing from the centre I, of the lower base of the cylinder, and 
through opposite edges of the upper base, those planes just 


SEGMENTS AND UNGULAS. 39 


pass along the sides, of the cone touching it on the two 
opposite sides, through its- whole length, and be cause the 
cone terminates in a point I in the centre of the lower 
base, there exists a portion FIDHC on one side of the 
cone, and a similar portion EIDGC on the other side, in- 
cluded between the surface of the cone, and that of the 
cylinder; and these two portions consist of regular pyramidals 
with cylindric bases and their vertices all centre in I, at the 
vertex of the cone ; for if lines be drawn from every point in 
their cylindric surface to the centre I, those lines will pass 
through every point in the solid portions to which those sur- 
faces belong. And because the cone DGCHL is supposed to be 
taken from the complement, these pyrimidals are elementary 
portions of a conesected cylinder, (Prop. XHI. Sch. B. UL. Hl. 
S. Geom. and Def. 2,). Hence (Prop. XII. Sch. 5, B. OL, £7. 
S. Geom.) they are equal to their cylindric bases multiplied by 
one-third of the radius IF. 


Cor. The opposite ungulas DEFA, CEFB and the com- 
plements IFDHC, IEDGC of aconesected cylinder are seg- 
ments which are in the same proportion to each other in their 
cylindric surfaces, as in their solidities. 


PROPOSITION V. THEOREM. 


The solidity of an ungula, whose base is less than half the base 
of the cylinder, is equal to its cylindric surface, multiplied by 
one-third of the radius of the base of the cylinder, minus the 
pyramidal segment of a cone, whose base is the base of the 
ungula, and whose vertex, is the point where the diagonal plune 
produced forming the ungula, would cut the axis of the cylin- 
der or its axis produced. 


Let BSHK be an ungula whose base SHK va 
is less than Jhalf the base of the cylinder, and 
its solidity will be equal to its cylindric sur- y 
face, multiplied by one-third of the radius of Le 
the base of the cylinder, minus the pyramidal ie 
segment KSHOof acone whose baseisKSH, JN [pp yy 
the base of the ungula ; and whose vertex | 
is O, the point where the plane BSH pro- 
duced to IE cuts the axis of the cylinder. 


My 


For the solidity of the ungula BFED, is equal to its auntie 
dric surface multiplied by one-third of the radius of the base 
of the cylinder (Prop. III,) and the section KSHFED minus 


40 SEGMENTS AND UNGULAS. 


the conical section KSHO is a portion of a coneseeted cylinder, 
which (Prop. XIU, B. Ill, Hi. S. Geom.) is equal to the product 
of its cylindric surface multiplied by one-third of the radius of 
the base, and because this section of the conesected cylinder 
is formed by a plane passing through the centre O, cutting it 
in such manner that every point in its cylindric surface may 
be connected by right lines with the vertice O, these lines 
being included in the same solid ; this section is also equal to 
its cylindric surface multiplied by one-third of the distance OD 
of this vertice from the cylindric surface; hence, the other 
two portions of the ungula, viz., the conical KSHO, and the 
ungula BSHK are equal to the remainder of the cylindric sur- 
face of the ungula BFED, viz., the cylindric surface of the 
small ungula BSHK multiplied by one-third of the radius OD 
of the base of the cylinder. Now, if from this product we 
take away the conical segment SHKO we shall have the ungula 
BSHK. Hence, &c. 


Scholium. If on the diagonal plane BSH of the ungula as 
a base, the pyramidal BSHP be described, P being the vertice 
of such pyramidal situated in the centre of the circular sec- 
tion of the cylinder, and in the plane KSH of the base of the 
ungula produced, the sum of this pyramidal and ungula is 
equal to their cylindric surface multiplied by one-third of the 
radius of the base of the cylinder ; and the pyramidal BSHP 
is equal to the conical portion, KSHO. 

For the pyramidal BSHP together with the ungula BSHK, 
constitutes the regular elementary pyramidal, BSHKP, with 
a cylindrical base, which (Prop. XIV. B. IIL, #2. S. Geom.) is 
equal to its cylindrical surface multiplied by one-third of 
the radius, PK. of the base, and because the sum of the 
ungula, BSHK, and pyramidal section of a cone KSHO, is 
equal to the same product (Prop. V.) it follows that the pyra- 
midal section BSHP is equal to the conical portion KSHO. 


PROPOSITION VI. THEOREM. 


The solidity of an ungula KLMD whose base is greater than 
half the base of the cylinder, is equal to its cylindrical sur- 
face multiplied by one-third of the radius of the base of the 
cylinder, plus a conical segment, whose base is the base of the 
ungula, and whose vertice is the point S, where the plane 
ELM cuts the azis of the cylinder. 


SEGMENTS AND UNGULAS. 4} 


Or the solidity of the ungulu is equal to the same product plus 
the pyramidal ELMO, whose base is the elliptical sections 
ELM and whose vertice is QO, the centre of the base of the 
cylinder. 


For through the point S pass the plane 
IF KC, and the ungula will be divided into 
two portions, one of which, EFCK, is equal 
(Prop. II.) to its cylindrical surface multi- 
plied by one third of the radius of the base 
of the cylinder; and the portion FCK 
LMD, from which, if we take a conical sec- 
tion LMDS, we shall have a portion of a 
conesected cylinder remaining which is equal 
to its cylindric surface multiplied by one third of its radius. 

Hence, the whole ungula ELMD is equal to its cylindric 
surface multiplied by one third of the radius of the base of the 
cylinder plus a segment of acone, whose base is the base 
of the ungula and vertice, the point where the plane ELM 
cuts the axis of the cylinder. 

Again, if we pass the two surfaces MOK and LOE meet- 
ing each other in the right line OH, those surfaces will 
cut out the pyramidal ELMO, whose base is the elliptical 
section ELM, and whose vertice is the centre of the base of 
the cylinder, which if we take from the ungula, will leave a 
portion which is equal to its cylindrical surface multiplied by 
one-third of the radius of the base of the cylinder, since it con- 
sists of a portion which may be divided indefinitely by 
planes passing through the vertice O, and each of those portions 
so divided will be perfect pyramidals, and will remain elemen- 
tary to the whole section. (Prop. XIII, Schls. B. Ill, £2 S. 
Geom.) Hence, &c. 

Cor. 1. Hence the pyramidal ELMO, is equal to the coni- 
cal segment LMDS. 

Cor. 2. As the plane section ELM is an ellipse, the surface of 
the pyramidal ELMO contiguous to the other portions of the 
ungula so divided is a conical surface ; for if the several 
points in the elliptical curve be connected with the vertice O, 
those lines must include a conical surface, since the section of 
a cone by a plane passing through both sides is an ellipse. 

Cor. 3. Ifthe ungula is so cut as to include the whole of 
the base BLDM, the circumstances will still be the same, and 
the ellipse of the section ELM becomes perfect, and the pyra- 
midal ELMO becomes a perfect oblique cone, and is equal to 
the conical body LMDS, which, in such case, becomes a per- 
fect right cone, whose base is the whole base BLDM of the 
cylinder, and whose altitude is half the altitude of the cylinder. 

4 


42 SEGMENTS AND UNGULAS. 


PROPOSITION YH. THEOREM. 


If in acylindric ungula ADB, whose base is equal to that 
of the cylinder, a cone be described on the same base and of an 
equal altitude with that of the ungula, the cone will be equal 
to two-thirds of the ungula; and each solid section of 
the cone made by planes parallel to the axis of the 
cylinder, and perpendicular to a plane passing through the 
axis of the cone and cylinder, will be equal to two-thirds of 
the corresponding section of the ungula cut off by the same 
plane. 


Let the oblique cone ABD eee 
of equal base and altitude to 
that of the ungula ADDNB, 
be supposed to be inscribed 
in the ungula, and the soli- 
dity of the cone will be 
equal to two-thirds of the 
ungula, and each solid sec- re! I P 
tion IRcD, PveD of the cone made by planes parallel to the 
axis of the cylinder, and perpendicular to a plane passing 
through the axis of the cone and cylinder, are equal to 2 their 
corresponding sections [RMbD, PuNiD of the ungula cut by 
the same planes. 

For, conceive an indefinite number of parallel planes 
IRM), NiPv, to be passed through the cone and ungula inde- 
finitely near to each other, and the two bodies will be divided 
into an indefinite number of strata, the sum of which, in each 
solid, will be in the relation of the magnitudes of the two 
bodies ; and the corresponding strata in each solid will be in- 
the relation of the superficial sections contiguous to such strata, 
in each, since by hypothesis the strata are indefinitely thin, so 
that no appreciable space intervenes between them ; hence, if 
the solidities of any stratum, or number of contiguous strata 
in each solid be represented by the superficial sections passing 
through such stratum or associated strata, they will exist in 
each in the relation of the superficial sections in each, made 
by the same planes. 

Now the superficial sections IRe, Pve &c., made by planes 
parallel to the side AB of the cone, are all parabolas, (by def.) 
and all the corresponding sections of the ungula are rec- 
tangles circumscribing the several parabolas, since each 
side of those sections Mb, Nr of the ungula are parallel to 
the opposite sides RI, vP, passing through the base. But 
the area of the parabola (Proposition VI, Book I,) is equal to 
two-thirds of its circumscribing rectangle or parallelogram, 
and as each of the several rectangles circumscribe their re- 


Ata te 


Ne | 


SEGMENTS AND UNGULAS, 43 
spective parabolas, throughout the whole extent of the two 
solids, it follows that not only is the cone equal to two-thirds 
of the ungula, but also that each solid section, or segment of 
the cone, cut by the planes RMOI, &c., is equal to two-thirds 
of the corresponding segment of the ungula cut by the same 
plane. 

Cor. Hence, as in Prop. VIII, Cor. B. Ill, El. S. Geom. a 
cone is equal to one-third of its circumscribing cylinder of the 
same base and altitude. 

Scholium. If the cone and ungula are cut by planes pa- 
rallel to the plane ADDN, the superficial sections of the ungula 
will be ellipses, or portions of ellipses, and the corresponding 
sections of the cone will be parabolas. 


PROPOSITION VIII. THEOREM. 


Two cylinders erected on the same base, and beiween the same 
parallel planes are equivalent or equal in solidity. 


Let ACDB and AEBD be two cylinders erected on the 
same base BD, and included between the same parallel planes 
BD and EAC, and the two cylinders will be equivalent. 

For let the cylinder ABDC be divided _,, ‘ 
into sections BDF f, FfgG by planes pa- 
rallel to the base, and let each of the sec- 
tions so divided, be removed from their 
positions in the cylinder ABDG, so that 
their several bases shall agree with, and 
become sections of the oblique cylinder 
AEBD ; viz., let the solid sections fF'Gg, 
be removed to the position Hhot, and let 
the other sections be similarly posited in 
reference to the two cylinders, so that the 
bases Hh, Mn, &c. of the several solid sections may be in- 
cluded in the oblique cylinder AEBD, and these several sec- 
tions will still be equal in their altitude to AB, the altitude of 
the cylinder from whence they are severally derived. Let, 
now, the number of these solid sections be indefinitelyfincreas- 
ed, and the altitude of each will be indefinitely small ; and 
hence the right cylinder will become identical with the oblique 
cylinder AEBD, and will still have the same altitude AB. 

Cor. 1. Since the base of the oblique cylinder AEBD is a 
circle, every section parallel to the base is likewise a circle, 
and equal to the base. 

Cor. 2. Hence every section of the oblique cylinder made 
by the planes perpendicular to its axis, is an ellipse. Therefore, 
if referred to its axis, it becomes an elliptical cylinder, but when 
referred to its base, it is circular. 


44 _ SEGMENTS AND UNGULAS. 


‘PROPOSITION IX. THEOREM. 


Cylindrical ungulas of the same base and equal altitude are 
equivalent. 


Let ABD and EBD, be two ungulas described on the same 
base BD ; and let their common altitude be AB, and the two 
ungulas will be equivalent, or equal in solidity. 

For each section FC or IL of the x 
two ungulas parallel to the base BD, 
have the same relation to the re- 
spective sections of their cylinders ; 
viz., the ungulical section FC has the 
same relation to the cylindric section 
Fh, as the ungulical section IL has 
to the cylindrical section IN ; and 
the equality of this relation remains through the whole alti- 
tude AB of the ungulas, and by Prop. VIII, it appears that 
two cylinders erected on the same base, and between the 
same parallels are equal in solidity. Therefore, the ungulas 
erected on the same base, and of the same altitude, are equi- 
valent. 

Scholium. The above proposition is manifestly true, whether 
the base of the ungulas is equal to that of the cylinder in 
which it is erected, or less than that base. 

Cor. 1. As the same arguments would apply to conical un- 
gulas, it may be inferred that conical ungulas of the same base 
and equal altitude are equivalent, if the cones from which 
they are taken are similar. 

Cor. 2. Since cylinders with equal bases are proportional 
to their altitudes, cylindrie ungulas on the same base are pro- 
portional to their altitudes. 


PROPOSITION X. THEOREM. 


Ifa cone be cut by a plane passing through the centre of the 
base, the solidity of the ungula formed by such plane will be 
equx! to its curve surface multiplied by one-third of its dis- 
tance from the centre of the base. 


Let the cone ABC be cut by the plane DEF, 
cutting off the ungula EFDB ; also by the 
plane EDG cutting off the ungula GEDB, or 
the ungula GEDF, and the solidities of the 
several ungulas will be equal to their curve 
surfaces multiplied by + of the distance LL of 
ce surface of the cone to the centre of the 

ase. 


SEGMENTS AND UNGULAS. eae 


For the sections may each be divided into an_ indifinite 
number of pyramidals or elements (Prop. XIII and XIV, B. 
Ill. Hi. S. Geom.) by dividing their curve surfaces as bases, 
and by passing the planes of division through the centre I of 
the cone’s base, at which point the several vertices of the py- 
ramidals all centre ; and because those pyramidals are all 
perfect; viz..as they include all the space intercepted between 
their bases and vertices in right lines, each one is equal to its 
base, multiplied by one-third of its altitude, and since the alti- 
tude is equal in each, and equal to the distance of the curve 
surface of the cone from the centre of its base, IL is that 
altitude, and hence IL is the common altitude of each of 
those pyramidals; hence the sum of the pyramidals con- 
stituting any section, is equal to the sum of their bases, con- 
stituting the curve surface of such section multiplied by one- 


third of IL. 


Cor. Hence, if a cone be cut by a plane passing threugh 
the centre of the base, the solidity of the cone, and its convex 
surface is divided in the same ratio, since the cone (Prop. [X. 
B. Ill., Hl. S. Geom.) is equal to its convex surface multiplied 
by 1 of its distance from the centre of the base. 


PROPOSITION XI. THEOREM. 


If a cone be described in a cylinder, and two parabolic ungulas 
be cut by planes passing through the centre of the cone’s base, 
the ungulas so cut, will be equal to two thirds of the unguli- 
cal complement of the cylinder, of equal base and altitude to 
the parabolical ungulas. 


Let BDS be a cone described in the cy- 
linder ABDC, and let EF Bd, and EF DM be 
two equal parabolic ungulas, described on 
and including the whole base of the cone, 
and the two ungulas will be equal to two- 
thirds of the ungulical complement BEDF-NP 
of the cylinder, of equal base and altitude. 


F 
For let the cone BDS be brought in the position BDA. and 
the parabolic ungula IEFD of the cone BDA will be equal to 
two-thirds of the semi-ungulical complement NPEFD of the 
cylinder, (Prop. VII ;) and because conical ungulas on the 
same base and equal altitudes are equivalent (Prop. IX Cor.,) 
the ungula IEF'D of the oblique cone BDA is equal to the un- 


46 SEGMENTS AND UNGULAS. 


gula EFDM of the right cone BDS; hence the parabolic un- 
gula EFMD is equal to two-thirds of the semi-ungulical com- 
plement NPEFD of the cylinder ; and the two similar ungu- 
las dEFB, MEFD are, together, equal to 3 of the ungulical 
complement BEDFPN of the cylinder. 
Cor. 1. Hence the conical complement 
EFdMS of the parabolic ungulas=the com- 
plement PMNd EF +the cone PMNds, is 
equal to two-thirds of the cylindrical ungula 
APNB. For if the whole cone BDA or 
BDS is=2 of the ungula BDA, (Prop. VIII,) 
and if the two conical ungulas dEFB, 
MEFD, as shown above, is = 2 of the com- 
plement P NBD, then must the complement 
EFdMS= 2 the cylindrical ungula APNB | 
Cor. 2. The small cone dMS, whose base passes through 
the vertices of the ungulas, is = to 4, of the cone BDS, since 
the diameter of its base is.necessarily = } that of the larger, 
and since they are similar solids ; for similar solids are to each 
other as the cubes of their like sides, (Prop. XXXV,B. Il, 
El. S. Geom.) and cube of 1 is 1, cube of 2 is 8 ; hence 178: : 
cone dMS: cone BDS. 
PROPOSITION XI, THEOREM. 
If a right cone be cut by a plane perpendicular to the plane of 
its base, the convex surface of the cone, and the plane of the 
base, will be divided in the same ratio by the cutting plane: 


Let DEF be a plane cutting the cone \ 
ABS, perpendicular through the base AEBF, = 
and the convex surface of the eone will be Ue 
divided by the cutting plane, in the same | 
ratio that the surface of the base is divided. [fe WA 

acy 


H F 

For since it is shown (Prop. XXII, B. Il, El S. Geom.,) 
that the convex surface of a pyramid and its base is divided 
in the same ratio, by a plane perpendicular to its base, and 
because a cone may be considered as a pyramid, whose cone 
vex surface consists of an indefinite number of planes, which 
are indefinitely narrow, it follows that if the convex surface 
and base of a right cone are cut by a plane perpendicular to 
the base, those surfaces will be each divided in the same ratio. 

Cor. 1. Hence if any portion of the base of a right cone is 
taken, the portion of the curve surface perpendicular above 
it, will be to the whole curve surface of the cone, as such 
portion of the base is to the whole base. 


SEGMENTS AND UNGULAS. AT 


Cor. 2. Hence, also, if a regular polygon, for instance a. 
square EF HG, be described in the base of the cone, and if on 
each side of this square, a plane be raised perpendicular to the 
base, the portion of the conical surface, cut off toward the 
axis, is to that of the rectilineal polygon EFHG, which cor- 
responds to it perpendicularly below, as the surface of the 
cone is to the area of its base; or as the slant side AS of the 
cone is tothe radius of the base; and, in fact, whatever 
figure be inscribed in the base, if we conceive a right prisma- 
tic surface raised perpendicular from the perimeter of the 
figure, it will cut off from the conical surface a portion which 
will be to it in the same ratio. 


Scholium. The solid sections DEFB, LGE, &c., are hyper- 
bolic ungulas, (Def. 7.) And if ungulas DEFB, LGE, &c., 
are taken from the cone, the remaining portion or compliment 
will be equal to its curve surface multiplied by.one-third of 
the distance of the curve surface of the cone from the centre 
of the base, + the surface of the plane hyperbolic sections ~ 
multiplied by one-third of their respective distance from the 
same centre of the base. 


Cor. Hence the portion of the cone included between the 
centre of the base, and that portion of the convex surface left 
by the ungulas, since it is equal to its convex surface, multi- 
plied by one-third of its perpendicular distance from the centre, 
the quadrature of which we have shown to be attainable, be- 
comes known in absolute terms, or its cubature is attained 
without regard. to the circle’s quadrature. 


PROPOSITION XIII. THEOREM. 


If an elliptical cylinder, and a circular cylinder, have equiva- 
lent bases and equal altitudes, they are equal in solidity ; and 
any ungulas similarly cut from each, with equivalent bases 
and ahitudes. are equivalent. 


For it has been shown (Prop. II. B. Ill. Zl. Geom.) that the 
solidity of a cylinder, with circular base, is equal to its base 
multiplied by its altitude ; and because this is true of any prism, 
whatever be the form of its base, (Prop. XVI. B. Il. El. 8. 
Geom.) it must be true of a cylinder with an elliptical base. 
Therefore, an elliptical and a circular cylinder of equivalent 
bases and equal altitudes, are equivalent. 


48 SEGMENTS AND UNGULAS. 


Again, let GEFB be an ungula cut from a 
circular cylinder ABDC; and if the base of 
this cylinder, including that of the ungula, is 
drawn out or elongated in the direction from 
BA toward DC, so as to become elliptical ; 
and if every section of the cylinder parallel 
to its base should become equal and similar 
ellipses, then the cylinder becomes an ellipti- 
cal cylinder. E 

Now, if planes should be passed through the cylinder par- 
allel to its axis, and in the direction AC or BD of its elonga- 
tion, in passing from a circular to an elliptical cylinder, this 
transformation will have been effected by an elongation of 
these planes proportional to the elliptical elongation of the 
base, or of the solid ; and, as this would be true in every par- 
allel plane, it follows that the elongation of those planes may 
be regarded as a measure of the ratio of enlargement of the 
solidity, by the same means. And as every such parallel sec- 
tion becomes enlarged in the same ratio, any specific portion 
of such section must suffer the same specific enlargement. 

And as the increase of any solid sections through which 
any portions of the plane sections pass, may be measured by 
the increment of those planes, it follows, that both the ungula 
and its complement are each increased in the ratio of the in- 
crements of the parallel sections passing through each in their 
enlargement. And the ungula which was cut from a circular 
cylinder, becomes the ungula of an elliptical cylinder, which 
ungula has become enlarged in the ratio of the enlargement of 
its base; and the solidity of the ungula from the elliptical cy- 
linder, is to the solidity of the ungula from the circular cylin- 
der, as the base of the former to the base of the latter. And the 
same would be true, if instead of an enlargement of the circu- 
lar cylinder to form the elliptical cylinder, it should be con- 
tracted in the direction AC, so as to give it eccentricity in the 
other direction ; but, by hypothesis, the cylinder and an ellipti- 
cal cylinder have equivalent bases; hence, ungulas similarly 
cut from each, are equivalent. 


Cor. 1. Hence, also, if a cone with a circular base, and one 
with an equivalent elliptical base, have equal altitudes, their 
solidities will be equivalent; and ungulas with equivalent 
bases and equal altitudes cut from each, are equivalent. 


Cor. 2. Ungulas, whose bases are the like parts of circular 
or elliptical cylinders, are as their altitudes; and it having 
been shown that they are also as their bases when their alti- 


- 


$< 


SEGMENTS AND UNGULAS. 49 


tudes are equal, it follows that they are generally as the rect- 
angle of their bases into their altitudes. 

Let ABEC and ABED be two 
ungulas cut from any cylinders, cir- 
cular or elliptical, such that AB 
shall be the same in each, and such 
that ID, the altitude of the base in | 
the first, shall be equal CE, the alti- D®auy 
titude of the second, and IC the al- 
titude of the base of the second shall be equal to the altitude 
ED of the first,-and the two ungulas so described will be 
equivalent. 


PROPOSITION XIV. THEOREM. 


The solidity of a cylindroid is equal to the product of the sum 
of the areas of the two ends, and four times the area of a pa- 
rallel section, equally distant between the two ends multiplied 
by 1 of the height. 


Demonstration same as for the prismoid, Prop. XXXV, 
Cor. B. Il, El. S. Geom., which see. 


PROPOSITION XV. THEOREM. 


If an ungula is cut from a sphere by two planes which inter- 
sect each other, not in the centre of the sphere, then the solidity of 
the ungula will be equal to its spherical base multiplied by one- 
third of the radius of the sphere, plus the products of the super- 

ficial sections of the ungula multiplied by one-third of the per- 
pendicular distances of the planes of those sections from the 
centre of the sphere, estimated from the sides of those planes op- 
posite the ungula. 


Let ABDE be an ungula cut by the 
planes HEI and FDG intersecting 
each other in the line AB, not passing 
through the centre of the sphere, then // 1 
will the solidity of the ungula ABDE be 4 ;* 
equal to the aiherteal aurtice ADBEA, [psy by —<al\ 
multiplied by one-third of the radius of H 
the sphere, — (the plane surface ABE, multiplied by one third 
of the distance CN) — (the plane surface ABD multiplied by 
one-third the distance CL) when those planes include the 
centre on opposite sides of each, from that of the ungula. 

For let each point in the perimeter, ABEA, ABDA, be con- 
nected by lines to the centre C, and the solid included by such 


f 
50 SEGMENTS AND UNGULAS. 


lines will be a spherical pyramid, whose solidity is equal to its 
spherical base multiplied by 2 the radius of the sphere; from 
which if we take away those portions included between the 
planes ABE, ABD and the centre C we shall have the ungula, 
But the pyramidal ABEC, is equal to the surface ABE. x 
CF, and the pyramidal ABDC is equal the surface ABD x 
CL. Hence as enunciated above. 

Scholium. If an ungula is cut by two 
planes which pass the centre before their 
intersection, so as to include the centre of 
the sphere within the ungula, then will its 
solidity be equal to the spherical surface 
multiplied by 4 the radius, plus the two py- 
ramidals erected on those plane sections, 
and whose vertices are in the centre of the 
sphere, which is agreeable to our proposition; for the sign be- 
comes changed from minus to plus according to the conditions. 

2. Also, if the two intersecting planes 
pass the centre on one side before their 
intersection, so as to cut out an oblique 
ungula ABDG, then since the pyramidal 
erected on the plane AFBP, and whose 
vertice is the centre, C, is considered ne- 
gative, and the pyramid erected on the 
plane GIDL is considered positive, by the proposition, then 
will the ungula be equal to its (spherical surface X + the ra- 
dius) — the plane APBF x (1 CN) + the plane GLDF 
x £CW 

3. And, generally, if the planes do not intersect each other 
within the sphere, the same proposition will still hold true, 
even though the planes may be parallel; in which case, the 
portion cut out will be a segment or zone of the sphere. 

PROPOSITION XVI. THEOREM. 


Lhe solidity of the second segment of a sphere is equal to its 
spherical surface, multiplied by 1 of the radius of its sphere 
minus each of the plane surfaces whose planes pass between 
the segment and centre of the sphere multiplied by + of their 
respective perpendicular distances from the centre, and plus 
each of the plane surfaces which include the centre of the 
sphere on the same side with the segment, multiplied by + the 
perpendicular distances of such planes to the centre: 

Let ABD be a segment of a sphere 
cut off by the plane ABCE, and if this 
segment is cut by the plane CEe perpen- 
dicular to the former plane, the two por- 
tions into which the segment is divided 
will be second segments. (Def. 6.) 


col com 


SEGMENTS AND UNGULAS. 51 


Draw AO, EO, CO, and eO, and the spherical pyramidal AE 
CeO will be equal to its spherical base x} AO or EO. Now; 
from this pyramidal may be taken the pyramidal whose base 
is the section CeEK, and whose altitude is KO, and also the 
pyramidal whose base is the section ACe and altitude HO, and 
there will remain the second segment AEeC. 

Also the second segment may be shown to consist of the 
sectoral pyramid, whose base is the spherical surface, plus a 
pyramidal CEeO, and minus the pyramid CeBO. Hence the 
proposition is true, as enunciated. 


PROPOSITION XVII. THEOREM. 


The solidity of a parabolie prism is equal to two-thirds of its 
circumscribing quadrangular prism. 


For the solidity of any prism: or solid, all of 
whose sections parallel to the base are equal to 
the base, is equal to the base multiplied by the 
altitude: and hence, prisms of the same alti- 
tude are as their bases ; but the base of the pa- 
rabolic prism, which is a parabola, is equal 
(Prop. VI, B. 1) 2 its circumscribing rectangle, 
also the rectangle circumscribing the base of 
the parabolic prism, is the base of the rectan- 
gular prism. 


PROPOSITION XVIII. THEOREM. 


Ifan ungula be cut from a parabolic prism by a plane pass- 
ing through the vertical line of the parabolic surface, and 
whose intersection forms an ordinate to the axis of the pa- 
rabola of the base; this ungula is equal to two-thirds of 
the ungulical complement of the prism of the same base and 
altitude. 


Let DECF be am ungula cut from the 
parabolic prism DEFCAB, by the plane 
CDE, from the vertical line CF of the pa- a 
rabolic surface. to the base DEF forming 
the ordinate ED to the axis FQ of the 
base, by its intersection with the plane of 
the base, and the ungula so cut will be 
equal to } of the ungulical complement 
DECAB. 

For, Jet a rectangle DELM be describ- D~ 
ed about the base of the ungula, and let any number of planes 


52 SEGMENTS ADN UNGULAS. 


NOf parallel to the base be passed through the ungula; and 
each of those sections made by those planes are parabolas ; 
and if about each of those parallel sections, rectangles NRPO 
are described, those parabolic sections will each be equal to 
two-thirds of their respective circumscribing rectangles, (Prop. 
VI. B. I.) Now, if these parallel sections through the un- 
gula are infinite in number and equidistant from each other, 
they will represent the whole ungula, and their sum may be 
taken as a function of the solidity of the ungula, and the sum 
of their several circumscribing rectangles may be taken as a 
similar function of the solid DECLMC; hence the ungula 
DEFC is equal to two-thirds of the solid CDELMC. Now, if 
we take the altitude FC of the ungula=QF, the axis of the 
parabola of the base, the side CLMC of the new solid will be 
a parabola equal and similar to the base DEF of the ungula, 
or ABC of the complement, since perpendiculars from every 
point in the perimeter of the base DEF, trace out the para- 
bolic section DEC ; and since perpendiculars from each point 
in the perimeter DNCOE to the plane CLM, includes and 
traces out the parabola CLM. And because QF or EL 
would be equal to EB, the rectangle DELM would be equal 
to the rectangle DEBA; hence, the two solids CABDEC, 
CLMDEC being similar figures on opposite sides of the same 
base, CDE are symmetrical, (Def. 19, B. Il. Hl. Sol. Geom.) ; 
and hence they are equivalent. .But the ungula CDEF has 
been shown to be equal to two-thirds of the solid CLMDEC, 
it is therefore equal to two-thirds of the ungulical complement 


CEDABC. 


Cor. 1. The above properties are true in Ea aon 


whatever part of the base the plane CDE may pias y 


cut, or whatever be the altitude of the ungula, 


provided the ungula and its complement have | f 
equal bases and altitudes; and because ungulas YZ Li 
on the same base are as their altitudes, an un- 

gula CGED is equal to the ungula GEDF, if fer 
the altitude CG=the altitude GF, since the two ungulas insist 
on or above the same base EDF. 


Cor. 2. Hence the solidity of the parabolic Pp G 
ungula CDEF is equal to 52. of its circumscrib- : 
ing prism, and the complement of a parabolic 
ungula is equal to 4 of its circumscribing prism. 
For the parabolic prism (Prop. XVII.) is equal 
to ; of its circumscribing rectangular prism ; 
and since the rectangular prism circumscrib- 
ing the parabolic prism, may be divided into 
two equal triangular prisms by a diagonal 


SEGMENTS AND UNGULAS. 53 


plane, identical with the plane which divides the ungula from 
its complement, it follows that the prism circumscribing the 
ungula DECF is equal to that circumscribing the complement 
DECI, equal half that circumscribing the parabolic prism. Let 
U equal the ungula, and C equal the complement, and let P 
equal the triangular prism circumscribing the ungula, and 2P 
will equal the quadrangular prism circumscriving the parabo- 
lic prism. Then will U+}C=4P and U=3C, hence C=13U. 
Substituting the value of C in the first equation, we have 
<1U=¢P-or 15U=8P. 

Hence U=P, 
or the ungula is equal to ;°; of its circumscribing prism. 

And if we substitute the value of U in terms of C in the 
first equation, we shall have 

12C=4P or 5C=4P. 

Hence C=4P, or the complement of the ungula is equal 
to 4 of its circumscribing triangular prism. 

Therefore, the portion CLBFE is equal to 75 of the prism 
CFNEBL. 

And the exterior portion CLEI is equal to } of the prism 
CANELLI. 


PROPOSITION XIX. THEOREM. 
* 


A parabolic pyramid or cone is equal to two-thirds of its cir- 
cumscribing rectangular pyramid. 


Let a pyramid be erected on a base whose figure is a para- 
bola, and if this base is circumscribed by a rectangle, it will 
=2 therectangle; and because every section of each of the 
solids erected on those figures as bases, and whose sections 
are in a common point are similar figures and similar to the 
base, the solids erected on those bases must be in the relation 
of their bases; hence the parabolic pyramid is =% its circum- 
scribing quadrangular pyramid. 


BOOK IIL. 


ON REVOLGIDS AND SOLIDS FORMED BY THE REVOLUTIONS OF THE 
CONIC SECTIONS. 


DEFINITIONS. 


1. A revotor is a solid generated by the continued semi- 
revolution of a polygon on axes parallel to the sides of the 
polygon respectively, and passing through its centre, which is 
fixed, and it includes all the space that is not cut off by either 
side of the plane, in their several semi-revolutions. 

2. Every revoloid has as many axes of rotation, as the 
polygon from which it is conceived to be generated, has inde- 
pendent sides ; but the axes of any two parallel sides coincide 
with each other and are identical. 

Thus, if the quadrilaterial ABDC be made to revolve on 
the two axes, EF and GH, parallel to its sides respectively, 
the solid generated hy the revolutions of those sides will 
be a revoloid, as represented by BCED. 


A G Kee 


B H +e 


8. A revoloid is designated by the number of sides con- 
tained in the figure from whence it is conceived to be gene- 
rated. . 

Thus a triangular revoloid is one formed by the continued 
semi-revolution of the sides of a triangle about their respective 
axes. | 

_A quadrangular revoloid, by the revolutions of the sides of 
a square about their corresponding axes. 


GN REVOLOIDS. 55 


Also a pentagonal, a hexagonal, or other polygonal revolord, 
is formed by the continued semi-revolution of the sides of a 
pentagon, a hecagon or other polygon, about their respective 
axes. | 


AB represents a hexagonal 
revoloid. 

4. The vertices of a revo- 
loid are its two extremities, 
where its several curve sur- 
faces meet at a point, as at A. 

5. The line joming these ex- 
tremities, is its transverse or 
vertical axis; and a section 
through this axis is a vertical 
section. 


WA 


Akt, G 
VAR) 


}, 


6. Any diameter perpendicular to the transverse axis, is a 
conjugate axis, and any section perpendicular to the trans- 
verse axis IS a Conjugate section. 

7. The figure from which a revoloid is supposed to be gene- 
rated, is called its prime; which is always represented by a 
conjugate section through the centre of the revoloid. 

8. A revoloid formed by the revolutions of the several sides 
of its prime on axes that are fixed or immoveable during the 
revolutions of each side respectively, may be called right re- 
voloids; and a vertical section through the centre of its op- 
posite sides is a circle. 

9. If, during the revelutions of the several sides of the prime 
of a revoloid, their axes are made to move in the line of the 
transverse or an opposite conjugate axis, then the revoloid 
will assume a different character according to the curve des- 
cribed by its sides. It may be elliptical, parabolic, or of any 
regular curve. | 

10. An elliptical revoloid is one whose transverse and con- 
jugate diameters are unequal, and whose vertical section 
through the centre of its sides is elliptical. 

11. A spheroid or ellipsoid is a solid generated by the revo- 
lution of a semi-ellipse about one of its axes, which remains 
fixed. 

If the ellipse revolve round the transverse 
or major axis AB, the figure is called a pro- 
late or oblong spheroid ; if the ellipse revolve 
round the shorter axis CD, the figure is called 
an oblate spheroid. 


12. A segment of a spheroid or of an elliptical revoloid is 


56 ON REVOLOIDS. 


& ‘ ° e 
a part cut off by a plane, parallel to the major or minor axis. 

13. A frustum of a spheroid and also of an elliptical revo- 
loid, is a part intercepted between two parallel planes, and is 
a portion included between two opposite segments. 

14. A parabolic revoloid is one whose vertical section through 
the centre of its sides, consists of two parabolas on opposite 
sides of the same base. 

15. A parobolic revoloid is vertical when the 
vertices of the several parobolic sections are 
identical with the vertices of the revoloid. This 
may also be called a parabolic pyramid. 


16. It is a conjugate parabolic revo- 
loid, when the vertices of the parabo- 
lic sections are all in a plane perpen- 
dicular to the vertical axis of the re- 
voloid. 


17. A Parabolic conoid is a solid form- 
ed by the revolution of a semi-parabola 
about its axis; it is also called a para- 
boloid. } 


18. A hyperbolic conoid, or a hyperboloid, 
is a solid formed by the revolution of a se- 
mi-hyperpola about its greater abscissa, or 
transverse axis produced. Thus, the hy- 
perbolic conoid MPBGN is formed by the 
revolution of the semi-hyperbola MPB, about 


through its sides consists of two hyperbolas whose vertices are 
in the plane of the conjugate axes. 

20. A hyperbolic pyramid or pyramoid, is one whose base 
is a polygon, and whose vertical sections through the vertice 
of the pyramoid, is an hyperbola. This may also be valled a 
vertical hyperbolic semi-revoloid. 


i 


ON REVOLOIDS. 57 


h , ; By ‘ 
21. A circular spindle is a solid generated by the revolution 
of a segment of a circle about its chord, which remains fixed. 
Cc 


22. An elliptical spindie is a 
solid generated by the revolution 
of the segment of an ellipse about 4 


its chord. “Sa at 


D 

23. A parabolic, or hyperbolic spindle, is a solid formed by 
the revolution of a segment of a parabola or hyperbola, about 
its ordinate. Thus, if the segment PBGP of Def. 18 be sup- 
posed to revolve about the ordinate PG, which remains fixed, 
it will describe a spindle. 

24. A revoloidal spindle is a revoloid 
circumscribing a spindle, a vertical section 
of which through the centre of its opposite 
sides, is equal and similar to that of the 
spindle through the same plane. It takes 
particular names according to the designa- 
tion of the inscribed spindle. 


25. A ringis a solid formed by the revolution of a plane sur- 
face about an axis exterior to itself, which axis is always in 
the same plane of the revolving surface. 

Thus, if the plane 0 


surfaces AB,CD, or EF “===? (=P 

are made to revolve E F 

about their several ax- 

es GH, IK, LM, they 

will severally describe © He! saglng M 
P 


solids, which are called rings. 

26. A ring is designated according to the figure of a conju- 

gate section, or of the plane surface from which it is generated. 

TT 

Thus from a rectilineal figure AB, is formed i, 
a prismatic ring. 


From the circle EF is formed a cylindrical ring. 

From the segment CD is formed an un- , 9 D 
gulical ring, which, if the line OP is equal to 
the radius of the circular area COD, it may 
be called a spherical ring, the curve surface 
forming a portion of the surface of a sphere. 
Rings formed from other figures or seg- 
ments of other curves, may be similarly designated. 

5 


§8 ON REVOLOIDS. 


' PROPOSITION I. THEOREM. 


A right revoloid is composed of cylindrical ungulas, equal in 
number to the sides of the revoloid; and these ungulas are 

* such as are formed by plane sections, from one side, meeting 
in the axis of the cylinder, the intersection of which planes 
forms a diameter to the cylinder. 


Cc if G D 


For, since the several curve sur- 
faces of a revoloid are conceived to 
be formed by the revolution of the 
sides of its prime about axes parallel 
to those sides respectively; (Def. 1,) 
the surfaces described by the revolu- 
tions of those sides, are cylindrical 
surfaces, (Def 1. B. Ill. El. S. Geom.) 
And since the angles formed by the 
sides of the prime are similar, each to 
each, in whatever similar part of their revolutions they may be, 
it follows that they would form similar angles with any plane 
perpendicular to these sides. Thus, because, when the side 
AB, of the revoloidal prime ABDC, in its revolution about its 
axis EF, comes into the position LL; and AC revolving about 
its axis GH, comes into the position II, &c.; the parts ce and 
ef of those sides form the same angle with each other as be- 
fore; they must also form the same angles with any planes 
AD or CB perpendicular to the plane ABDC or cefg. Hence, 
the figure cefg is similar to the prime ABDC; and because 
this is true in whatever similar positions, the several sides of 
the prime may be in the course of their revolutions, it follows, 
therefore, that sections through the angles meeting in the 
transverse axis of the revoloid, are plane angles ; and as these 
plane sections cut the cylindric surfaces diagonally through 
the axis of the revoloid, the segments so cut are cylindric un- 
gulas, (Def. 3. B. IT.) 

Hence, each side of a right revoloid is a cylindric ungula, 
such as is formed by plane sections cutting the cylinder diag- 
onally, and meeting in the axis of the cylinder and forming a 
diameter thereto. 


Cor. 1. Hence any vertical section through the angles 
formed by the meeting of the curve surfaces of the revoloid, 
or any vertical section of the revoloid not at right angles to the 
sides, is an ellipse. 


Cor, 2. Hence, also, the elliptical revoloid may be con- 
ceived to be made up of similar ungulas cut from an elliptical 


ON REVOLOIDS. 59 


cylinder ; see Prop. XUI. B. IL, and a parabolic revoloid may 
consist of ungulas from a parabolic cylinder or prism. 


Cor. 3. Since a revoloid is composed of cylindrical ungu-. 
las equal in number to the number of the sides, those ungulas 
are also such as are-cut from the cylinder of altitudes or lengths 
on the cylindric surface equal to the lengths of the sides of the 
revoloid indicated by the lengths of the sides of its prime. 


Cor. 4. All sections of a revoloid perpendicular to the ver- 
tical or transverse axis, are similar figures, since the sides of 
its prime retain their parallel position in whatever part of their 
revolution they are supposed to be taken. 


PROPOSITION H. THEOREM. 


The solidity of every right revoloid, bears the same relation to 
that of its greatest inscribed sphere, as the area of its prime 
does to that of its greatest inscribed circle. 


For since the surface of the revoloid is composed of cylin- 
‘dric surfaces, (Prop. 1.) and since vertical sections of a right 
revoloid through the centres of its opposite sides are circles, 
(Def. 8.) it follows that a sphere may be inscribed in the re- 
voloid so as to touch its cylindric surfaces through the whole 
circumference of those circular sections, viz., through the cen- 
tres of the vertical sides, so that vertical sections through the 
centres of the sides in the revoloid, will be identical with those 
of the inscribed sphere through the same planes. 

Now, if planes be passed through the two solids perpendi- 
cular to the vertical or transverse axis, those planes will cut 
the solids in the relation of their solidities through such sec- 
tions; and as all such sections (Prop. I. Cor. 4.) are similar 
figures and similar to its prime, and as all the corresponding 
sections of the sphere are similar figures, viz., circles; and be- 
cause the inscribed sphere touches the surface of the revoloid 
through their whole vertical sections; each of the conjugate 
sections of the revoloid are polygons, circumscribing the cor- 
responding circular sections of the sphere. And because an 
indeflinite number of parallel plane sections may be regarded 
as a function of the solidities of the bodies through which such 
sections pass, the sum of a series of the revoloidal sections, 
equidistant and parallel to each other, is to the sum of a similar 
series of sections of the sphere, as the solidity of the revoloid 
to the solidity of the sphere; hence the solidity of the revoloid 
is to the solidity of the sphere as the area of its prime to the 
area of its inscribed circle. 


60 ON REVOLOIDS. 


Cor. Hence, also, by a parity of reasoning in reference to 
the perimeters of all the similar polygons formed by the con- 
jugate sections of the revoloid which circumscribe and touch 
all their corresponding circumferences of the sections of the 
inscribed sphere ; the whole surface of the revoloid is to that 
of the inscribed sphere as the perimeter of the polygon of the 
revoloidal section to the circumference of its inscribed circle. j 


Scholium. The proposition and corollary are also manifestly 
true in reference to any revoloid, and its inscribed solid of_re- 
volution, if the surfaces of the two solids correspond through 
vertical sections through the centres of the revoloidal sides. 


PROPOSITION III. THEOREM. 


The area of each facial side of a right revoloid is equal to that 
of the corresponding side of its circumscribing prism. 


Let ABDA be a vertical hemisphere of a right quadrangu- 
lar revoloid, and CD its axis; and let AEIFB be a prism cir- 
cumscribing the hemisphere; then will each facial side of the 
hemisphere of the revoloid be equal to its corresponding sur- 
face of the prism. 

For first, let a re- 

ular semi-polygon 
AGLNO, &c. be des- 
cribed about the fig- 
ure representing the 
revoloid, and from 
the angles draw the 
lines GH, LM, &c., 
and if this is conceiv- 
ed to be done on 
every facial side of the revoloid included within such lines as 
the boundaries of surfaces, it would be a polyedron, whose seve- 
ral faces AGHB, GLMH, LNOM, &c., correspond with, and 
include the several faces of the revoloid. Now, since by hy- 
pothesis the polygon AGLN, &c., circumscribing the figure, 
is regular the distance across the face of the polyedron, in the 
plane of a vertical section, is equal in each, viz., TV the width 
of the face, GLMH is equal VD the width of the face LNOM, 
and they are each equal to GL or LN; although we cannot so 
represent those spaces in the diagram in consequence of the 
curvature in a plane perpendicular to this sheet. But it is evi- 
dent that the sides GL, IN of the polygon will truly represent 
the breadth of those faces. And since those several faces are 


ON REVOLOIDS. 61 


trapezium, they are severally equal to half the sum of their 
parallel sides multiplied by their altitudes or width. Thus the 
face MH is equal to (Q(GH+iLM) xX LG or VT, and since VT 
is supposed equal to LG; and since IK is equal 1GH+1iLM, 
IK x LG equal the face GLMH ; the face LVOM=PQx LN, 
and the face AGHB=ABXAG. Now the areas of the cor- 
responding parallel spaces or sections in the prism being par- 
allelograms, are severally equal to their lengths multiplied by 
their breadths. Thus the space GRSH corresponding to the 
face GLMH, is equal GH XGR, and the parallelogram REFS 
corresponding to the face LNOM is equal RSX RE. 

Now let IK be produced each way to W, X and PQ to Y, Z; 
and through the point I draw the radial lines Ca Cb, draw also 
Id perpendicular to AC. Then in the right-angled triangles 
LRG, alG having an accute angle at G common, they are simi- 
Jar, (Prop. XXU, B. IV. £1. Geom.), and hence their sides are 
proportional ; and because the right-angled triangle CAa has 
an accute angle at a common with the right-angled triangle 
Gla, this triangle is also similar to the former, as also the tri- 
angle Cdl or UIC. Hence GR: GL:: UI: CI, and because 
IK=2UI, and AB or GH=2CI, we have GR: GL:: IK: GH. 
That is, RG which is a factor of the parallelogram GRSH is 
to GL, which is the factor of the trapezium GLMH as IK, 
another factor of GLMH to GH another factor of GRSH; 
whence, by multiplymg extremes and means, we have 
GRxGH=GLXxXIK, viz., the surface of the face GLMH of 
the polyedron is equal to the corresponding vertical surface 
GRSH of the prism; and since the same may be shown in 
reference to any other of the faces with its corresponding por- 
tion of the surface of the prism, it follows that the whole sum 
of the polyedral faces on one side, is equal to the whole cor 
responding surface of the prism. Let the number of the pol- 
yedral faces be indefinitely increased, and the truth of the pro- 
position is still manifest, but when the faces are indefinitely in- 
creased, they become assimilated to that of the body about 
which they are described; therefore, the facial surface of 
each side of a right quadrangular revoloid is equal to that of 
the corresponding side of its circumscribing prism. 


Cor. 1. Let us suppose the revoloid, instead of being quad- 
rangular, to consist of any number of facial sides; then, by 
hypothesis, its circumscribing prism will consist of an equal 
number of vertical sides and in the same ratio each to each; 
and hence the proposition is true for a revoloid of any number 
of sides. Let, then, the number of sides be indefinitely in- 
creased, the reveloid then becomes a sphere, and the circum- 


62 ON REVOLOIDS. 


scribing prism becomes a cylinder. Hence the surface of a 
sphere is equal to the convex. surface of its circumseribing. 
cylinder. 


Cor. 2. Hence the surface of a revoloid is equal to the pe- 
rimeter of its centeral conjugate section multiplied by the ver- 
tical axis, for the vertical surfaces of the circumscribing prisny 
are equal to this product; and since in a right quadrangular 
revoloid its circumscribing prism has six equal sides, four 
of which are equal to the surface of the revoloid, it follows 
that the whole surface of the revoloid is to the whole surface 
of the prism as 2 to 3. ¢ 

If |) represent the perimeter of the central conjugate section, 
D the vertical axis. or diameter, then }D will represent the’ 
surface, and if be the altitude of any zone by sections par- 


allel to the conjugate axis, then will H=the curve surface of 


the zone or segment_of the revoloid or sphere. 


Cor. 3. Hence also the surface of a sphere is equal to its 
circumference multiplied by its diameter or altitude, for the 
curve surface of its circumscribing cylinder ig equal to this 
product. And since the surface of a great circle of the sphere 
is measured by the product of its circumference into half the 
radius, or by 1 the diameter, (Prop. XVI. B. V. El. Geom.) 
Therefore the surface of a sphere is four times the area of its 
great circle: this is equal to 4rR’, (Prop. XIII, Sch. 3. HZ. S. 
Geom.) and because the two bases of the eircumscribing cylin- 
der are each equal to one of those eircles, it follows that the 
whole surface of the cylinder is equal to six of those circles, 
and hence that the whole surface of the sphere is to the whole 
surface of the cylinder as 2 to 3, as before found in the ele- 
ments of solid geometry. . 


Cor. 4. Since we have shown that the surface GLMH of 
the polyedron is equal to the surface GRSH contained within 
the same paralle! planes, it follows that the surface of any zone 
or segment LNOM either of a revoloid or sphere, is equal to 
the perimeter or circumference of a central conjugate section 
multiplied by the altitude of such zone or segment. 


Cor. 5. The surface of two zones taken in the same revo- : 


loid or sphere, or in equal revoloids or spheres, are to each 
other as the altitudes; and the surface of any zone, is to the 
surface of the sphere, as the altitude or diameter of the zone is 


to that of the sphere. Hence the surfaces of every parallel. 


portion of equal altitude are equal. 


| 


ON REVOLOIDS. 63° 


PROPOSITION IV. THEOREM. 


— Lf a cylindrical ungula be cut by two planes from the same side 
of the cylinder, the intersection of which planes forms a di- 
ameter to the cylinder, and if the altitude of the ungula, or the 
extreme length of the ungula taken in the direction parallel to 
the axis of the cylinder, is equal to the cylinder’s eircumfer- 
ence, then the sections or ungulas so cut, will be equal to a 
sphere described in the cylinder, or to a sphere whose diameter 
as equal to that of the cylinder. 


Let ABCD be a cylinder, and let HK be a sphere of equal 
_ diameter described in the cylinder, and let AJBbe an ungula cut 
from the cylinder by the two planes OLJ, MNJ, from the points 
A and B, whose distance ‘AB parallel to the axis UT is equal 
to the circumference of the cylinder, and let the cutting planes 
meet in J forming a diameter to the cylinder; then will the | 
section AJB be equal to the sphere HK. 


For, let planes be passed through the ungula and sphere 
perpendicular to the diameter formed by the intersection of the 
planes LOJ and NMJ; and these plane sections, formed by 
these planes in each solid, will be proportional to the magni- 
tudes of the solid through such sections. Now, any section 
of the ungula, by a plane perpendicular to the diameter which 
passes through the pole J of the sphere, is a triangle; anda 
section through the sphere made by the same plane, is a circle; 
and the area of a triangle formed by any section, is equal to its 
base multiplied by half its altitude on such base. Thus the 
area of the triangle JAB=ABX2H4J; the area of a parallel sec- 
tion Jcc, is in like manner = the base ccX}# the altitude J3; and 
so for the area of any other parallel section Jbb, Jaa, &c. ; and 
because these triangles are equiangular, (Prop. XXII. B. IV. 
il. Geom.) their bases are proportional to their altitudes ; thus, 
HJ: AB::J3:cc::J2:6b::J1: aa, &c. The areas of the 


r 
ee ON REVOLOIDS. 


several circles formed by the same planes passing through the 
sphere, are equal to their circumferences multiplied by 4 their 
several radii; thus the area of the great circle of the sphere’ 
HXKY is equal to the circumference HXKY X?2 the radius 
HJ, the area of the circle PQ corresponding to the section Jcc, 
is equal to the circumference PQ X 3 radius J P, and the areas of 
the circles RS, &c., = their several circumferences multi- 
plied by 1 their several radii. And because the circumfer- 
ence of circles are to each other as their radii, (Prop. XV. B. 
V. El. Geom.) as radius JK : circle HK: : radius J3 : circle 
PQ : : radius J2 : circumference RS, &c.; and, since this is the 
ratio of the lines AB, cc, bb, &c., as shown above, the several 
circumferences are in the same ratio of those lines as bases of 
their several triangles ; but the line AB by hypothesis, is equal 
to the circumference HK ; hence the several circumferences 
PQ, RS, 1J, &c., are respectively equal to the several bases 
cc, bb, aa, &c.; hence, also, the areas of the several circles be- 
ing sections of the sphere, are respectively equal to their seve- 
ral corresponding triangles, being sections of the ungula made 
by the same planes; and as this is true whatever may be the 
number of the parallel sections, or in whatever position they 
are taken, it follows that the solidity of the ungula is equal to 
that of the sphere. 


Cor. 1. Since the section AJB may be regarded as com- 
posed of the two unglas AJH, BJH, regarding JH as their com- 
mon base, and because ungulas of the same base are propor- 
tional to their altitudes, it follows that if we cut the ungula 
HgJ, whose altitude Hg=1 HB, and the ungula HJf, whose 
altitude Hf=4 HA, then the section fJg including those un- 
gulas together with an equal opposite section, Jz are equal to 
the section AJB, consequently equal to the sphere HK. Or if 
we take the section CTD, whose base CD=AB, this section 
will also be equal to the sphere HK. 


Cor. 2. Since, it may be shown, that all sections of a 
spheroid or ellipsoid, by planes parallel to its axis of revolu- 
tion, are similar ellipses; and since ellipses are to their inscrib- 
ed circles, as the diameter'of the circle to the major axis of the 
ellipse, (Proposition IX of the Ellipse, B. I,) the solidity of an 
ellipsoid HWKYV is to the solidity of the sphere HYKX, as 
the axis of revolution VW of the ellipsoid is to the axis of re- 
volution XY of the sphere, and hence if an ungula, whose base 
is equal to half the base of the cylinder be taken, and whose 
altitude is to that of the ungula CTD as the axis of revolution 
VW of the ellipsoid to the axis of revolution XY of the sphere, 
that ungula will be equal to the ellipsoid HWKYV made by the 
revolution of the ellipse on its axis VW. 


e 


ON REVOLOIDS. , 65 


Cor. 3. As the ellipsoid HWKY is to the sphere HYKX 

as the axis VW of the ellipsoid to the axis XY of the sphere, 
* and as the cylinder fhig circumscribing the ellipsoid is to the 

cylinder, pgrs circumscribing the sphere, as the same axis VW 
to the same axis XY, or as the length of those cylinders res- 
respectively ; the ellipsoid HWKV bears the same ratio to its 
circumscribing cylinder fhig, as the sphere HYKX to its cir- 
cumscribing cylinder pgrs. IfS=the sphere, and E=the el- 
lipsoid, and if P =the cylinder circumscribing the ellipsoid, and 
Q=the cylinder circumscribing the sphere, 

then, XY:S::VW:E 

and, ea: Qs 3 VW Pe 

hence, (Prop. XIX. B. I. Hl. Geom.) 8: Q::E: P. 


Scholium.. The segment AJB being equal to the sphere HK, 
the annexed figure may represent the manner in which they 


are convertible into each other, as there exists no mathema- 
tical reason why the segment AJB may not be changed, as 
partly represented in the figure. 


66 ON REVOLOIDS. 


Cor. 4. It is also evident that an ungula 
BCDH, whose base BCD is the segment of a 
circle, similar to a segment AHB of a vertical 
section through this circular spindle AHBG ; 
if the altitude DH of the ungula is equal to the 
circumference of the conjugate section HG of 
the spindle, the solidity of the ungula will be 
equal to that of the spindle, so also will its 
curve surface. | 


AB shall be equal to the circumfer- 
ence of a circle of which HG is the 
radius, and AG and BG lie in the 
planes ACE, BFD, then may the 
segment be converted into a ring whose outside diameter is 
equal 2HG; and every section of the ring will be equal to a 
section through the segment. 


Cor, 5. Hence, also, if a cylindric segment ABC be cut by 
two planes meeting in the surface of the cylinder oT C, and 
A D 


C 
terminating at A and B on the opposite 
side, the distance of which points from 
each other is equal to the circumference f 
of a circle whose radius is CD; the seg- EM 
ment so cut may be changed in the form 
represented by EF, which is the form of 


C 

and at such an angle that the distance FG shall he equal to the 
circumference of a circle whose radius is CD, and the section 
AFGB will form a cylindric, ring, whose inner diameter is 
equal to twice CE. 


ON REVOLOIDS. 67 


Cor. 6. And because the section ANF, (see diagram above,) 
if cut from its position and placed in the position BGP, com- 
pletes the cylinder NFPB, which is 
equal in length to half the sum of sides 
FG and AB. The solidity of a cylin- 
dric ring AB, is equal to that of a cy- 
linder whose base is equal to a radial 
section of the ring, and whose altitude 
is equal to half the sum of its inner ana 
outer circles. Andhence, cylindricrings 
whose sections are equal, are proportional to their inner or 
outer circumferences. 


PROPOSITION V. THEOREM. 


The solidity of a cylinder circumscribing a sphere, is equal to 
the solidity of a prism circumscribing the cylindrical ungula 
vr ungulas whose solidity 1s equal to the sphere. 


Let MNGL be a cylinder circumscribing the sphere PK; 


and let CRDBKA be a prism circumscribing the two similar 
ungulas QRKV, SRKYV described on the base, KRV=half 


the base of the cylinder from which they are conceived to be 
taken ; and if QS, the sum of their altitudes, is equal in length 
to the circumference of the cylinder, then (Prop. IV.) will the 
ungulas equal the sphere of equal diameter to that of the cy- 
linder, and the cylinder equal MNLG will be equal to the 
prism CRDBKA. 

For the altitude AC or KR of the prism, on the base CRD, 
is equal to the altitude of the cylinder; GL being=the axis, both 
of the cylinder and the sphere; and the length of the side CD 
of the prism is equal to the sum of the altitudes VQ, VS, or 
the length QS of the ungulas taken on the surface of the cy- 
linder ; but the length QS is equal to the circumference of 


foal 
68 | ON REVOLOIDS. 


the sphere or cylinder by hypothesis. Now, the area of the 
base LN of the cylinder, is equal to its circumference mul- 
tiplied by half the radius RE; and the area of the base CDR 
of the prism, is equal to the line CD, or the circumference of 
the cylinder multiplied by half the line RE; hence, the base of 
the prism is equal to that of the cylinder. And the solidity 
of the cylinder is equal to its base ENL multiplied by its al- 
titude GL; the solidity of the prism is also equal to its base 
CDR, or the base of the cylinder multiplied by its altitude 
RK; hence, the solidity of the cylinder is equal to that of the 
prism. 


Cor. As the sum of two or more ungulas of equal base, are 
equal to one greater of the same base, if the sum of their alti- 
tudes is equal to the altitude of the greater, (Pr. 1X.Cor. 2. B. IL.) 
and because a prism circumscribing an ungula is proportional 
to the altitude of the ungula, two or more prisms circumscrib- 
ing ungulas, the sum of which is equal to a sphere, are equal 
to the cylinder circumscribing that sphere; and also, the se- 
veral prisms circumscribing ungulas equal to a revoloid, are 
equal to the prism circumscribing the revoloid composed of 
those ungulas. 


PROPOSITION. VI. THEOREM. 


Every right revoloid is equal to two-thirds its circumscribing 
prism ; and every sphere is equal to two-thirds its circum- 
scribing cylinder. 

Let ABCD, (fig. 1.) be a revoloid, or a sphere circumscribed 
by the prisin or cylinder EFGH ; then will the revoloid ABCD 
be equal to two-thirds the prism EFGH ; and the sphere ABCD 
will be equal to two thirds of the cylinder EFGH. 

For let the plane EG be a Fig. 1. 
vertical section through the » A 
centre of the revoloid and 
prism bisecting their oppo- 
site sides ; in which position, 
(Def. 8.) the section of the 
revoloid is a circle, and the 
section of the prism through 
the same plane is evidently a 
rectangle. These sections 
are, also, evidently those of a 
sphere and its circumscribing 
cylinder, made by a plane 
through the common. axis, 


AC, of the sphere and cylin- 


H C G 


_" 


“~ 
1% SS, : 
» 


ON REVOLOIDS. l 69 


der, (Prop. 1, Sph. Geom.) and (Def. 5, B. 1,) through the centre, 
I, join EI, FI; also let AIC, as an axis, be parallel to EH or 
FG; and DIB and KL parallel to EF, or HG, the base of the 
section of the prism or cylinder, the latter line, KL, meeting 
FI in M, and the circular section of the revoloid or sphere in 
N; and the plane EIE will represent the vertical section of a 
pyramid of equal base to that of the prism, and an altitude, IA : 
or it will represent a vertical section of a cone of equal base 
to that of the cylinder and of an altitude, IA. 

Now, if the line KL produced, if nececsary, be conceived 
to revolve on the axis AC, it will cut conjugate sections of those 
solids in the relation of their magnitudes, viz: KS the section 
of a prism or cylinder, KN the section of a revoloid or sphere, 
and KM the section of a pyramid or cone. 

Now, AF being equal to Al or IB, and KL parallel to AF, 
then by similar triangles IK = KM, (Prop. XVII, B. IV, £7. 
Geom.,) and since, in the right angled triangle, IRN, IN’=1K? 
+KN?, (Prop. XXIV, B. IV, Ll. Geom.) and, because, KL is 
equal to the radius IB or IN, and KM=IK, therefore, KL?= 
KM?+KN?’; or the longest line forming the section of the 
prism or cylinder, is equal to the sum of the squares of the 
other two, forming sections of the revoloid or sphere, the py- 
ramid or cone. 

Let now the conjugate sec- 
tions of those solids formed by 
the revolution of the lines KL, 
be represented. (Fig. 2.) 

Thus let the square OALB, re- 
present the conjugate section of 
a prism, described about the 
quadrangular revoloid, and let 
the circle OALB be the section 
of a cylinder circumscribing 
the sphere ; the square PCND 
will represent the section of 
the the revoloid, and the circle 
PCND will represent the sec- 
tion of the sphere ; the square SEMF will represent the sec- 
tion of the pyramid, and the circle SEMF a section of the 
cone. Now, since we have shown that KL’, or the square 
AL described on the line KL is equal to KN’ + KM”, or the 
square CN + the square EM, being squares described on the 
lines KN and KM; it results that the square OALB, described 
on the line OL, which is double the line KL, is equal to the 
sum of the squares PCND, SEMF, described on the lines 
PN, SM, being double the lines KN, KM. (Prop. XII, B. I, 


70 ON REVOLOIDS. 


El. Geom.) That is, the section of the prism is equal to both 
the sections of the revoloid and pyramid. And because cir- 
cles described on the diameters SM, PN, OL, are proportional 
to the squares described on those diameters, respectively, 
(Prop. XIV, B. V, Hl. Geom.) it follows that the circle OA LB, 
is equal to the two circles PCND, SEMF'; that is, the section 
of the.cylinder is equal to the two sections of the sphere and 
cone. And because, (Prop. I, Cor. 4,) the sections of a revo- 
loid, of its circumscribing prism, its inscribed pyramid of any 
number of sides, are similar figures; they are, (Prop. XIV, 
Cor. 3, B. V, Ll. Geom.) proportional as the squares of the 
diameters of their inscribed circles, hence the section of a 
prism of any number of sides circumscribing a revoloid, is 
equal to the sum of the corresponding sections of the revoloid 
and pyramid, and as this is always true in any parallel position 
of the revolving line KL, (Fig. 1.) it follows, that the prism, 
EB, circumscribing the hemisphere DAB of the revoloid, being 
composed of all the former sections is equal to the hemisphere 
DAB of the revoloid and pyramid EIF, composed of all the 
latter sections ; and that the cylinder EB, is, in. like manner, 
equal to the hemisphere DAB and coneIEF. But the pyramid 
IEF is a third part of the prism DEFB; (Prop. XX VI, Cor. 1, 
B. I, Hl. S. Geom.) consequently, the hemisphere DAB of the 
revoloid, is equal to the remaining two-thirds. And the cone 
IEF, (Prop. VIII, Cor. 1, B. Ill, 7. S. Geom.) is equal to one- 
third of the cylinder, DEFB: hence, the hemisphere DAB, 
is equal to the remaining two-thirds of the cylinder. 


Cor. 1. A pyramid, revoloid, and prism, are to each other 
as the numbers 1, 2, and 3, when the bases of the pyramid 
and prism are each equal to the prime of the revoloid, and 
when their altitudes are all equal. Also, a cone, sphere, and 
cylinder, are, in like manner, proportional as the numbers 1, 
2, and 3, if the base of the cone and cylinder are each equal 
to the great circle of the sphere, and if the altitude of the 
cone and cylinder are each equal to the diameter of the sphere. 


Cor. 2. All spheres, and all similar revoloids are to each 
other as the cubes of their diameters, all being like parts of 
their circumscribing cylinders or prisms. 


Cor 3. From the foregoing demonstration, it appears, that 
the revoloidal frustum DBNP, is equal to the difference be- 
tween the prism DBLO, and the pyramid SIM, all of the same 
common height IK. And that the revoloidal segment, PAN, 
is equal to the difference between the prism, EF LO, and the 


ON REVOLOIDS. 71 


pyramidal frustum EFMS, all of the same common alti- 
tude AK. 

And the same is trite of any parallel segment or frustum of 
the sphere, cylinder and cone. 


Cor. 4. The sphere may be regarded as a revoloid whose 
prime has an infinite number of sides, which (Prop. XII, Cor. 
4, B. V, El. Geom.) is identical with a circle. 


Cor. 5. From the above de- 
monstration, and from propo- 
sition fourth, it may be shown 
that any ungula, being a por- 
tion of the revoloid, is equal to 
two-thirds of its circumscribing 
prism or wedge. ‘Thus, if the 
ungula ABDC be such as forms 
a portion of a revoloid, it will ; 
be equal to two-thirds of its circumscribing prism, BGFEA, 
since this ungula is the same part of the revoloid as its cir- 
cumscribing prism, is of the prism, circumscribing the revo- 
loid. 


Hence, if from the cylinder EF any ungulas GCH, ACB, 
GCJ, GCa, &c., are taken, whose cutting planes meet in the 
centre, C, of the cylinder or anywhere in its axis, those ungu- 
las are each equal to two-thirds of their circumscribing prism 


or wedge; these being ungulas such as may compose a right 
revoloid, and (Prop. IX, Cor. 2, B. II.) they are proportional to 
their altitudes or lengths taken on the surface of the cylinder ; 
viz: the ungula GH, which may be considered as composed 
of the two ungulas GCJ, HCJ; their common base being the 
section through the line CJ, is to the ungula ACB, as the line 
CH to the line AB, &c. Anda prism circumscribing an un- 


: 
7 


72 ON REVOLOIDS. 


gula is evidently proportional to the length of those lines, forrn- 
ing one of the dimensions of one of its sides. 


Cor. 6. Since all the vertical sections of a right revoloid 
except those bisecting its sides at right angles, are ellipses, 
(Prop. I, Cor. 1.) it may be inferred that an elliptical revoloid 
is also equal to two-thirds of its circumscribing prism ; and 
hence that an ellipsoid is also equal to two-thirds its circum- 
scribing cylinder. For the ellipsoid evidently bears the same 
relation to the elliptical revoloid as the sphere does to the right 
revoloid, since the elliptical revoloid may, by multiplying the 
number of the sides of its prime, be shaded off into the ellip- 
soid without changing its relation to its circumscribing prism, 
which in such case becomes a cylinder. 


Scholium. Lest this latter corollary should not appear suffi- 
ciently satisfactory in view of the important principle enunci- 
ated; and as it may seem to require a more rigorous demon- 
stration, it will be made the subject of the following propo- 
sition. 


PROPOSITION VIL. THEOREM. 


An elliptical revoloid is equal to two thirds its circumscribing 
prism; and an ellipsoid is equal to two-thirds of its circum- 
scribing cylinder. 


Let us imagine the annexed 
figure ABCD, to be a conju- 
gate section of a right quad- 
rangular revoloid by a plane 
through its centre, and. let 
abcd, be regarded as another 
conjugate section of the re- 
voloid between the central 
section and the extremity of 
the axis,and let them be so 
projected that the axis I, 
perpendicular to the plane 
of projection may be sup- | 

osed to pass through their centres; and let the circles 

FGH, and efgh be similar sections of a sphere of the same 
diameter and similarly posited. 

Now let it be conceived that vertical sections or ungulas 
be taken from each side of the” revoloid and sphere ; such 
that KIL, All, shall be sections of the ungula from the side 


ON REVOLOIDS. 73 


AIB, then will the sectoral portions of those triangles be respect- 
ively, sections through the spherical ungulas; now let the plane 
sections of the revoloid and sphere be reduced so as to close 
the spaces formed by the triangles so removed, but retaining 
their former figures. Thus let ABCD be reduced to opgqr ; 
abdc to stuv, &c., then, because (Prop. XXIII., B. 1, £7. 
Geom.) if any number of squares are proportional, the sides of 
those squares are proportional, it follows that if P be = the 
length of the side AB of the revoloidal section, and P’ = the 
length of the sides or, and if p = the length of the side ab 
and p’ = the length of the side st ; then will P? = the square 
ABDCand P”=the samesquare reduced, or opqr ; p? will=the 
square abdc and p” = the same square reduced, or stuv. 

And we have shown that P’: P?:: p?: p” 

Hence also (Prop. XX, B. I, El. Geom.) P: P’::p ae) 

That is, as the side AB is to the same side reduced, so is the 
side ab, to the same side reduced, and hence the sides, or dia- 
meters of those sections are reduced by the removal of the 
ungulas, in the ratio of those sides, or their diameters respect- 
ively, and because the circumferences of circles are as their 
diameters, and their areas as their squares, the diameters of 
the circular sections of the sphere, are each reduced in the 
ratio of these diameters ; and the same will hold true with 
regard to any parallel sections of the revoloid or sphere. 

Now the several radii El, el, &c. of the several sections 
may be regarded as ordinates to the vertical axis, I, of the re- 
voloid and sphere ; andthe radii ml, wl, &c., may be regarded 
as the corresponding ordinates of the vertical sections of the 
solids so reduced. Now, since these latter ordinates are seve- 
rally proportional to their corresponding ordinates of a vertical 
circular section of the revoloid and sphere, it follows that the 
same sections made by the same plane, through the solids so 
reduced are ellipses, for (Prop. XII, Cor. Hillipse) this is a 
property of an ellipse, when compared with a circle ; hence 
the revoloid becomes an elliptical revoloid (Def. 10) and the 
sphere becomes an ellipsoid. 

Now if a prism is supposed to circumscribe the revoloid 
before being reduced, and if a cylinder is supposed to circum- 
scribe the sphere, they must in order to accommodate them- 
selves to the elliptical revoloid and ellipsoid, be reduced in 
every conjugate section, equal in amount to the reduction of 
the central conjugate sections of the revoloid and sphere ; and 
hence the prism will have been reduced in the same ratio as 
that of the revoloid; and the cylinder will in like manner, have 
been reduced in the same ratio as that of the sphere, so that if 
(Prop. V1) a right revoloid is equal to 3 of its circumscribing 


6 


74 ON REVOLOIDS. 


prism, and a sphere is equai to 2 of its circumscribing cylin- 
der, then also will an elliptical revoloid be equal to 2 of its cir- 
cumscribing prism, and an ellipsoid will be equal to 2 of its 
circumscribing cylinder. 


Cor. Since an elliptical revoloid is formed of ungulas, cut 
from an elliptical cylinder (Prop. I, Cor. 2) whose bases 
are severally the semi-base of the cylinder, it follows that such 
ungulas ofan elliptical cylinder, are equal to 2 of their respec- 
tive circumscribing prisms. 


Scholium. Since the solidity of a cylinder may be expressed 
by *R?X H (Prop. I, Sch. B. Ill., Hl S. Geom.,) that is 
since its solidity is as the square of its radius or diameter 
multiplied by its height, it follows that ellipsoids are propor- 
tional to each other as the square of their revolving axes mul- 
tiplied by their fixed axes, and hence the same is true also of 
the elliptical revoloid, If R?*H = the solidity of a cylinder, 

2 R2arH = the solidity of a spheroid 

the solidity of a prism will be 4R*H, 

and the revoloid will be £ R?H, 

or if D = 2R then the revoloid is = 2 D?H. 


Cor. 2. As in Cor. 8, prop. VI, in relation to the segments 
of a right revoloid, or a sphere, so in relation to the segments 
of an elliptical revoloid, or spheroid, they are respectively 
equal to the difference between the corresponding sections of 
their circumscribing prisms or cylinders, and inscribed pyra- 
mids or cones. (See diagrams Prop. VI). 


Cor. 3. Since a spheroid is equivalent to a sphere drawn out 
as in the case of a prolate, or contracted, as in the case of 
an oblate spheroid, and in such manner, as that every line or 
section through the spheroid, in the direction of the expansion 
or contraction, is drawn out or contracted, in the ratio of the 
increase or decrease of the axis in such direction, it follows that 
any segment of aspheroid, by a plane parallel to its axis of re- 
volution, is to a corresponding segment of its inscribed sphere, 
ifa prolate spheroid, or that of its circumscribing sphere if 
the spheroid is oblate, as the diameter of that sphere to the 
axis of the spheroid, or as in the conjugate axis of the seg- 
ment’s base to the transverse of the base, or the axis parallel 
to the axis of the spheroid. 

And any segment by a plane perpendicular to the axis is 
also proportional to a corresponding segment of the sphere 
from which it may be conceived to be produced as the axis of 
the sphere to the axis of the spheroid, or as the height of a 
similar spherical segment to the height or altitude of the seg- 
ment of the spheroid. 


ON REVOLOIDS. 75 


— Scholium. Let ABCD be i E 
the complement of acylinder ~;EERBERyEE= 
from which is taken the un- 
gula GEFH=a quadrant of a 
revoloid, and also a similar WANA 
opposite ungula, cut by l a aT 
planes meeting in the diame- Ml ! ll 
ter EF; and there may be | | 
taken two cones whose bases 
are the two bases of the com- ell Lee 
plement, and whose vertices D | EF. C 
are in the centre J, andthe 

parts remaining will be equal to the remaining cylindrical sur- 
face X 1 the radius of the base or distance JE’. (Prop. IV., 
B. I.) 

Let the complement be divid- 4 
ed in the line EF, and let the seg- 
ments be inverted so that the 
bases shall comprehend the line 
EF, and if the planes ABJ, DCJ 
cut off the segments AEBJ, 
DFCJ then there shall be left the 
pyramid AJ,BJ,CJ,DJ,whose base 
is equal to a central section of the 
cylinder along its axis; viz. 
ABCD, and its vertical heightis | 
equal to the radius of the cylin- D 
der ; and as each side of the cy- 
linder is supposed to be cut alike, we shall have two of those 
pyramids, which together are equal to one-third of the prism 
circumscribing the revoloid. 

It follows therefore that the two ungulas together with the 
two pyramids are equal to a full quadrangular revoloid. 


== i 
— a 
i ait ee 
ce 


il 
th 


Hence there remains four portions ABJFH to be determined, 
which when placed together, so that their several vertices J, 
shall coincide, their cylindric surfaces turned inward, their 
plane surfaces will be outward, forming a pyramid equal to 


76 ON REVOLOIDS. 


one of the former pyramids, minus a pyramidal portion 
PSQRJ, which is required to complete the pyramid. It will 
be perceived that every section pqrs of this latter solid, pa- 
rallel to the base is a square, and = the square of pq, the 
versed sine of the arc Jp, therefore this solid is equal to the 
squares of an infinite series of equi- J 

distant versed sines drawn into 
their distance; or is to its circum- 
scribed prism, erected on the same 
base PSQR as an infinite series of 
the squares of equidistant versed :s 
sines to a similar series of the : 
squares of radii, as will be more fully discussed in another place. 


PROPOSITION VIII. THEOREM. 


If the solidity of a sphere is equal to one or several cylindrical 
ungulas of the same cylinder, the surface of the sphere will 
also be equal to the cylindrical surface of such ungula or 
ungulas. 


Let HK bea sphere AJB a cylindric ungula equal to the 
sphere, then will the surface of the sphere be equal to the 
cylindrical surface of the ungula. 


D 


For let an indefinite number of planes be passed through the 
two solids perpendicular to the axis J of the sphere, formed by 
the intersection of the planes LOJ, MNJ, and the sphere will be 
divided into an indefinite number of circles, from the great 
circle of the sphere down to the smallest about the axis, and 
the ungula will be divided in like manner into an indefinite’ 
number of similar triangles, with bases AB, cc, bb, &c., 
which was shown (Prop. IV) to be equal to the circumfer- 
ence of the circles through the corresponding sections of the 
sphere ; and because this is the case throughout, it follows 


‘ON REVOLOIDS. 77 


that the surface of the sphere, which may be represented by 
the sum of the several circumferences of the circles, is equal to 
the cylindric surface of the ungula, which may likewise be 
represented by the sum of the several bases of the triangles. 
And because cylindric ungulas of the same base are pro- 
portional as their altitudes, (Prop. IX, Cor. 2, B. II) and be- 
cause their cylindrical surfaces are proportional to their alti- 
tudes, it follows that if several ungulas, cut from a cylinder of 
a given diameter are equal to a sphere of the same diameter, the 
surface of the sphere will be equal to the sum of their cylindri- 
cal surfaces. | 


PROPOSITION IX. THEOREM. 


The solidity of a sphere as well as a revoloid, is equal to the 
product of its surface by one-third of its radius. 


For since the revoloid ts made upofsections of the cylinder, 
whose several solidities are equal to their curve surface mul- 
tiplied by one-third of the radius of the cylinder, whence 
they are conceived to be taken (Prop. III, B. If) which radius 
is equal to the vertical height of the several elimentary pyra- 
mids of which these sections are formed, and is also the 
radius of the revoloid, it follows that the solidity of the whole 
revoloid composed of all the sections, is equal to the whole 
curve surface of all the sections multiplied by one-third the 
radius. 

Thus, if the revoloid consist of six facial sides a,b,c,d,e,f, the 
solidity of each of which is equal to its surface x + radius 

oe i 
or ys or surface a X tr = solidity a, surface b X 3r = solli- 
dity b, &c., their surface a+b+c+d+e+f X fr equal to the 
solidity of the whole revoloid, and as the number of sides of 
a revoloid may be increased indefinitely without altering the 
relation of its elimentary pyramids, it follows that the same 
relation exists between the solidity of the sphere, and its sur- 
face, as in the revoloid, viz., the solidity of a sphere, as also of 
a revoloid, is equal to the product of their respective surfaces by 
one-third of their respective radii. 


Scholium. 1. Conceive also a polyedron, all of whose faces. 
touch the sphere; this polyedron may be considerd as formed of 
pyramids, each having for its vertex the centre of the sphere, 
and for its base one of the polyedrons faces. Now, it is evident 
that all these pyramids will have the radius of the sphere for 
their common altitude ; so that each pyramid will be equal to 


78 ON REVOLOIDS. 


one face of the polyedron multiplied by one-third of the radius ; 
hence the whole polyedron will be equal to its whole surface 
multiplied by a third of the radius of the inscribed sphere (Prop. 
XV. Cor. B. Il. Hl. S. Geom.) It is therefore manifest that 
the solidities of polyedrons, as well as revoloids circumscribed 
about a sphere, are to each other as the surfaces of those polye- 
drons or revoloids respectively. 

Now, also, as with a revoloid, the number of polyedron’s faces 
may be inscribed till the polydron becomes identical with the 
sphere, and then its solidity is equal to the product of its surface 
with one-third of its radius ; hence the sphere may be conceiv- 
ed to be made up of an indefinite number of indefinitely small 
pyramidals, whose bases when associated, form the surface of 
the sphere, and this surface has the same relation to the whole 
solid, or the sphere, asthe base of each individual pyramidal 
has to the solidity of each. 


Cor. 1. Hence the solidity of any sector of asphere or of a 
revoloid is equal to its spherical or cylindrical surface multi- 
plied by } of the radius, fora sector consists of an association 
of regular pyramidals, the sum of whose bases form the curve 
surface of the sector. 


Scholium. 2. Since the axis of the cylinder circumscribing 
a sphere is equal to its diameter, its solidity is equal to its whole 
surface, including the two ends multiplied by a third of the ra- 
dius. For it may be conceived to be made up of two cones, 
whose bases are two ends of the cylinder, and vertical height 
= half the axis or length of the cylinder = radius ; and the ele- 
mentary pyramids of the curve surface, whose vertices termi- 
inate in the centre of the cylinder with those of the cones, and 
hence, its solidity bears the same ratio toits surface, that the 
solidity of a sphere, a right revoloid, or polyedron, circumscrib- 
ing a sphere do to their respective surfaces. 


Cor. 2. Hence we have three orders of surfaces, which, 
taken as bases of pyramids, and multiplied by one-third of the 
distance of such base to the vertice of the pyramid, will de- 
termine the solidity of such pyramid; but, as observed in 
Schol. to Prop. III, B. II, #7. 8. Geom. in reference to cylindri- 
cal surfaces, the vertice of the pyramid with a spherical base, 
must be in the centre of the spherical curvature. 


Cor. 3. Hence, as the solidity of a sphere is equal to two- 
thirds that of its circumscribing cylinder, and as the surface 
of the sphere is equal to the curve surface of the cylinder ; 
and as the solidities of each of these bodies are equal to the 


ON REVOLOIDS. 79 


products of their respective surfaces, by one-third of their com- 
mon radius, the surface of the two ends of the cylinder is 
equal to half the curve surface, for if @ = the surface of the 
sphere, and z = the two bases of the cylinder, then will a + 
x = the whole surface of the cylinder, including the ends. 
Then aX +r or radius = the solidity of the sphere, and $ ra + 
1yz = the solidily of the cylinder. 

Butt ra=Gra+ ire) xX 2=¢ra+ ire. 

Transposing and dividing } a = 2 rz. 

Hence a = 2z, therefore the area of the two ends is equal 
to half the area of the convex surface of the cylinder. 

The same may be inferred from the ratio of the inscribed 
cones, to the remaining portion of the cylinders. 


Scholium. 3. Since the surface of a sphere whose radius is 
R, is expressed by 47R? (Prop. III, Cor. 2,) it follows that the 
surfaces of spheres are to each other as the squares of their 
radii; and since their solidities are as their surfaces multiplied 
by their radii, it follows that the solidities of spheres are to 
each other as the cubes of their radii or diameters, and the 
same is true also of revoloids. If the diameter is called D, 
we shall have R = } D, and R* = 1D* ; hence the solidity of 
we sphere may likewise be expressed by 4 7 X } D?= 

xD)’. 


PROPOSITION X. THEOREM. 


Every segment of a sphere is measured by the half sum of its 
bases multiplied by its altitude, plus the solidity of a sphere 
whose diameter is this same altitude. 


Let BH, DL, be the radii of the two bases 
of a segment, HL its altitude, the segment be- 
ing generated by the revolution of the circular 
zone DLHB, about the axis AG passing 
through the centre of curvature C ; from C 
draw CO perpendicular to the chord DB, 
draw also the radii CD,CB. The solid de- 
scribed by the section BCD is measured by 
2”, CB’, LH (Prop. XXIII, Sch. 2, B. IIL, 
El. S. Geom.) ; but the solid described by , 
the isosceles triangle DCB, has for its mea- 
sure 2” .CO.LH,(Prop. XVII, Cor. B. III, HZ. S. Geom..,) ; 
hence the solid described by the segment BDO = 2.7. LH. 


80 ON REVOLOIDS. 


(CB’—CO’). Now, in the right angled triangle CBO, we have 
CB’ — CO? = BO* = } BD?’ ; hence the solid described by the 
segment BDO, will have for its mearsure 27. LH .} BD’, 
or} « BD? LH. 

Again, the solid described by the trapezeium BDLH is 
= 1i¢LH.(BH*—DL’+BH.DL,) (Prop. X, B. III, EZ S. 
Geom.) Hence the segment of the sphere, which is the sum of 
those two solids, must be equal to} 7. HL, 2@BH+2DL+2 
BH. DL+BK’.) But since BK is parallel to HL, we have 
DK = DL — BH, hence DK? = DL?—2DL.BH + BH? 
(Prop. IX, B. IV., Hl. Geom.) ; and consequently BD? = BK’ 
+DK? = HL’?+ DL’? — 2DL.BH+ AH’. Substitute this 
value for BD? in the expression for the segment, omitting the 
parts which cancel each other, we shall obtain for the solidity 
of the segment } rHL. (83BH’ + 3DL?+HL’), an expression 
which may be decomposed into two parts; the one 3 +. HL, 


(3BH?-+3DH?), or HL. CR tea 


the bases X by the altitude ; while the other } + . HL’ repre- 
sents the sphere of which HL is the diameter. The same 
may be proved of any other segment DE, EF’, &c. ; hence the 
proposition is manifest. 


being the half sum of 


Cor. 1. Ifeither of the bases is nothing, the segment in ques- 
tion becomes a spherical segment, with a single base ; hence 
any spherical segment with a single base is equivalent to half 
the cylinder, having the same base and altitude + the sphere 
of which this altitude is the diameter. 


Cor. 2. If IC is perpendicular to AC the solid described by 
the revolution of the segment about the axis AC, isa ring, 
when if DC or BC be the radius of curvature of the arc DB 
of the segment, it is (Def. 26) a spherical ring, and in such case 
the ring is equivalent to a sphere, whose diameter is the alti- 
tude of the segment from which the ring is taken. 


Cor. 3. We may hence infer that every segment of a right 
revoloid included between two parallel planes perpendicular 
to its transverse axis, is also measured by the half sum of its 
bases multiplied by its altitude, plus the solidity of a similar 
revoloid, whose diameter is this same altitude ; and that the 
solidity of the belt ABCD, (see diagram to next proposition,) 
taken from the middle segment or zone of a revoloid, is equi- 
valent to a right revoloid of similar type, whose diameter is 
this same altitude. 


ON REVOLOIDS. 81 


Cor. 4. It may be inferred also that every segment of an 
elliptical revoloid, or of anellipsoid, cut by planes perpendicu- 
lar to their vertical or transverse axis, is equal to half the 
sum of its bases multiplied by its altitude + the solidity of a 
similar revoloid or spheroid, whose vertical axis is this same 
altitude, since the spheroid is the sphere drawn out as in the 
prolate, and constructed as in the oblate spheroid, and since 
the same is true of the elliptical revoloid, compared with the 
right revoloid ; and hence that the solidity of a belt taken from 
the middlezone or segment of an elliptical revoloid is equivalent 
to a similar revoloid, whose vertical axis is the altitude of the 
segment or breadth of the belt, and the same is also true of the 
ring taken from the middle segment of a spheroid, viz., that it 
is equivalent to a similar spheroid, whose axis of revolution is 
the altitude of the segment or breadth of the ring. 


PROPOSITION XI. THEOREM. 


The ungular portions of a revoloidal belt taken from the mid- 
dle zone of a quadrangular revoloid, are together equal to a 
revoloidal spindle, whose verticle section through its opposite 
sides is a double segment of a circle, equal to the segment con- 
stituting a conjugate section of the belt. : 


Let ABCD be a belt taken from the middle zone of a quad- 
rangular revoloid, and let its conjugate section IES or KES 
be the segment of a circle ; and the angular portions KAIE, 
LBMF, NCOG, and PDEQH will be equal to the revoloidal 
spindle RUT V whose vertical section through the centre of its 
opposite sides is a similar double segment of the same or an 
equal circle. 

For, let the prismatic cylin- Af i 
dric ungulas IESLF, FMNG, x 
OGPH, HQ KES, be removed, = = 
and let these several angular 
spaces be brought in juxta-posi- ij 
tion by the contact of their cor- = | 
responding faces, thus let FLM | 
be brought in the positiort IES, so 
that LEFT shall coincide with IES, 
let. GNO and HQP be brought in 


contact with FML and EIK, so 971 1A“ A= 
that GN, HQ, shall coin- P O 


cide with FM, EK. For since the faces of these angular 
portions are all similar and equal to each other, they would co- 
incide when placed in contact each to each, and because the 


82 ON REVOLOIDS. 


corresponding faces are parallel to each other in the belt, they 
would, when placed in contact, form the same angular space as 
in the belt, and because they form all the angular space about 
the axis EI when in contact they will form a body, whose con- 
jugate section is a square, and whose vertical section through 
the centre of its sides is the double segment of a circle, which 
(Def. 24) is a revoloidal spindle. 


Cor. 1. Hence, we may infer that the angular portions of 
any revoloidal belt taken from the middle frustum or zone 
of a revoloid of any number of sides, is equal to a revoloidal 
spindle of the same number of sides, and whose vertical sec- 
tion through its opposite sides is equal to a double segment of 
the circle composing a section of the belt. And sinee the an- 
gular space is the same in a circular ring as ina circle or about 
a point, (Prop. XXI. B. V. El. Geom.) it follows that the an- 
gular space of a belt or ring taken from the middle zone of a 
sphere, is equal to a circular spindle formed by the revolution 
of a section of the ring, being a circular segment about its 
chord as an axis. 


Cor. 2. Therefore, a belt from the middle zone of a revo- 
loid of any number of sides, may be resolved into prismatic 
ungulas equal in number to the number of the sides of the re- 
voloid, the bases of which are severally equal to segments of 
the circle forming a conjugate section of the belt, and whose 
altitudes are each equal] to the length of the sides of the poly- 
gon forming the mner portion of the belt ; and one perfect re- 
voloidal spindle whose vertical section through the centre of 
its opposite sides, is the double segment of the circle, each 
equal to the segment formed by a section of the belt. 

Hence, the solidity of a revoloidal belt is equal to that of a 
prismatic cylindrical ungula, whose base is the section of the 
belt and whose altitude is equal to the perimeter of the inner 
surface of the belt, plus a revoloidal spindle whose vertical 
section through its opposite sides.is the double segment of the 
section of the belt. , me 


Cor. 3. The solidity of a ring taken from the middle zone 
of a sphere, is equivalent to that of a prismatic cylindrical un- 
gula, whose base is the section of the ring, and whose altitude 
is equal to the inner circumference of the ring, plus a circular 
spindle formed by the revolution of the segment, forming a 
section of the ring about its chord as an axis. 


_ Scholium 1. Let Z = the middle zone of a revoloid, P = 


ON REVOLOIDS. 83 


the prism formed by taking a belt from this zone; then will 
Z— P = the belt. 

Let S = the segment formed by a section of the belt, and 
p = the perimeter of the inner surface of the belt, and S = 
the solidity of the spindle formed by the angular portions of 
the belt ; then will sxp or sp = the prismatic cylindrical un- 
gulas, and Z — P — sp = s = the revoloidal spindle. 


Scholium 2. Let Z = the middle frustum of a sphere, P = 
the cylinder after taking away the ring whose section is the 
segment of the circle of which the spherical surface of the 
segment is formed, and Z — P = the ring. 

Let S = the segment formed by a section of the ring and 
x = the inner circumference of the ring, and sx will = the 
prismatic cylindrical ungula which is conceived to form a por- 
tion of the ring. Let s = the spindle formed from the angu- 
lar portions, and Z— P— sx = s = the circular spindle. 


Scholium 3. Let the length or perimeter of a revolidal belt or 
spherical ring, taken from the middle zone of a revoloid or 
sphere, be increased while the conjugate section remains the 
same, and the portion which may be resolved into the prisma- 
tic cylindrical ungula is increased in the same ratio: but the 
angular portion which we resolve into the spindle, remains 
constant. 

Therefore, rings formed from portions, whose sections are a 
similar segment of a circle, are not proportional to the circum- 
ferences of the rings, unless the chord of the segment is per- 
pendicular to the axis of rotation of the segment generating 


the rings. (See Cor. 6th. Prop. IV.) 


Cor. 4. If a belt or rg be taken from a middle frustum or 
zone of an elliptical revoloid or an ellipsoid, the belt or ring 
may, in like manner, be resolved into a prismatic ungula of 
an elliptical cylinder whose base is the section of the belt or 
ring, and whose altitude is its inner circumference ; and an 
elliptical spindle, such as would be formed by the revolution of 
the elliptical segment, formed by a section of the belt or ring 
around its chord as an axis, which spindle is equal to all the 
angula space contained in the belt or ring. 


Scholium. 4. The same remarks as were used in reference 
to segments Prop. X., Cor. 4, are applicable to the belt 
or ring from a middle zone or frustum of an elliptical revoloid 
or an ellipsoid, viz: that a belt from a middle zone of an ellip- 
tical revoloid is equivalent to a similar revoloid whose axis 


84 ON REVOLOIDS. 


is equal to the altitude of that of the zone, and that a ring 
from the middle zone of an ellipsoid is equivalent to a similar 
ellipsoid whose axis is equal to the altitude of the segment ; 
and this is true whether it is an oblate or prolate revoloid or 
ellipsoid. 


Scholium. The middle frustum of a circular or an elliptical 
spindle, may be resolved into two portions ; a cylinder whose 
base is one of the bases of the frustum, and its altitude equal 
to the length of the frustum ; and a ring remaining after this 
cylinder is withdrawn. The ring may be computed by find- 
ing what portion it is of a spherical ring, and the cylinder 
may be computed according to its dimensions. 


PROPOSITION XII. THEOREM. 


A vertical parabolic revoloid is equal to half its circumscribing 
prism; anda vertical paraboloid is equal to half its curcum- 
scribing cylinder. 


Let ABC be a vertical section of a A i 
segment of a quadrangular parabolic a 


revoloid, or of a paraboloid, and let fea 
BCHI be a similar section of the prism iin TS. 
circumscribing the revoloid, or cylin- 1a RA 
der circumscribing the paraboloid 

through the same plane HBADIC, *® D e 
and if we suppose the axis AD to be 

divided into an indefinite number of 

equal parts through which, and through the solid, if planes 
EFG are passed perpendicular to such axis, the sections of 
the revoloid made by such planes would be squares, all of 
which make up the revoloid, and the sections of the paraboloid 
would be circles, all of which constitute the paraboloid, which, 
as those squares and circular sections are indefinitely near 
together, may be represented as a function of those squares or 
circles. 

Now, because the square described on FG is equal to} of 
the square on EG, 4FG?= the area of the square described on 
HG.  Butby property of parabola, (Prop. VII., parabola,) 
pxAF = FG? where p denotes the parameter of the pa- 
rabola ; consequently 4px AF will also express the same 
square section EG, and therefore 4px the sum of all the 
AF’s will be the sum ofall the square sections, or the same func- 
tion of the whole content of the revoloid ; and because circles 
are as the squares described on their diameters, the sum of 
all the AF’sX willbe a similar function of the whole content of 
the paraboloid. 


: ON REVOLOIDS. 85 


But all the AF’s form an arithmetical progression, begin- 
ning at 0 or nothing, and having the greater term, and the 
sum of all the terms, each expressed by the whole axis AD. 

And since the sum of all the terms of such progression is 
equal to AD, multiplied by AD, or } AD2, half the product 
of the greatest term, and the number of terms; therefore 
z AD? is equal to the sum of the AF’s, and consequently 
4p X}AD?, or 2XpXAD? is the sum of the revoloid. But by 
the properties of the parabola p: DC:: DC: AD, 

C23 


orp= 7p 3 consequently 2xpxAD?, becomes 2X AD 


x DC? for the solid content of the revoloid. But4xAD 
x DC? is equal to the prism HIBC, consequently the parabo- 
lic revoloid is equal to half of its circumscribing prism, and 
from a parity of reasoning, with regard to the paraboloid and 
cylinder, the paraboloid is equal to half its circumscribing 
cylinder. 

Cor. Hence each of the ungulas of which the parabolic re- 
voloid is composed is equal to half its circumscribing prism, 
and these ungulas are such as are cut from a parabolic cylin- 
der or prism, by planes meeting in the vertical plane, passing 
through the vertices of its two parabolic bases. 


PROPOSITION XIII. THEOREM. 


The solidity of a frustum of a polar hemisphere of a parabolic 
revoloid is equal to a prism of equal altitude, and whose base 
is half the sum of the two bases of the frustum ; and the 
solidity of a frustum of a polar hemisphere of a paraboloid 
is equal to that of a cylinder of equal altitude, and whose 
base is equal to half the sum of the two circular bases of the 
frustum. 


For in the frustum BEGC last proposition. 
2p x AD? = the solid ABC 
and 2px AF? = the solid AEG, 

Therefore the difference 2px (AD? — AF?) = the frustum 
BEGC, 

But AD?—AF2= DFx(AD+AF,) 
therefore, 2px DF x (AD+AF) = the frustum BEGC. 

But, by the parabola px AD = DC’, and px AF = FG’, 
therefore 2x DF x (DC’?+FG’) = the frustum BEGC, 
that is frustum BEGC = half the sum DC, FG, of the frus- 
tum multiplied by the altitude DF. 

And for the same reasons as adduced in the the last pro- 
position, this demonstration is equally applicable to the frustum 
of a revoloid or a paraboloid. | 


/ 


£6 ON REVOLOIDS. 


Cor. Hence any conjugate section of an ungula, being part 
of a polar hemisphere of a parabolic revoloid, is equal to a 
prism of equal altitude, and whose base is equal to half the 
sum of the two bases. | 


PROPOSITION XIV. THEOREM. 


A conjugate parabolic revoloid is equal to ; of its circumscrib- 
ing prism; and a conjugate parobolic spindle is equal to 
zs of its circumscribing cylinder. 


Let AIBC bea parabolic revoloid 
whose vertical axis is an ordinate 
to the several parabolic § sections, 
and whose vertices are all in the 
plane made by the conjugate sec- 
tion ARPB, and the revoloid will be 
equal to 3% of its circumscribing 
prism. 


For this revoloid is composed of parabolic ungulas, such as 
are cut from the vertical side of a parabolical prism, which 
(Prop. XVIII. Cor. 2, B IL.) are equal to 3; their circumscribing 
prisms ; hence a number of associated ungulas are equal to ;% 
their associated prisms, and since the inscribed paraboloid 
bears the same proportion to its circumscribing cylinder, as 
the revoloid to its circumscribing prism; hence a conjugate 
paraboloid ora parabolic spindle is 3 of its circumscribing 
prism. 


Scholium. The frustum of a parabolic spindle may be re- 
solved into three portions—first, a cylinder whose base is one 
ofthe bases of the frustum, and whose altitude is the length of 
the frustum—second, the angular portion of the ring re- 
maining after taking away the cylinder, which is equivalent 
to a parabolic spindle formed by the revolution of the section 
of the ring on the chord or double ordinate—third, a parabo- 
lic prism, whose base is a section of the ring, and whose alti- 
titude is equal to the inner circumference of the ring. 

The first is equal to its base multiplied by its altitude. The 
second is equal to 3’; of its circumscribing prism ;the third is 
equal to 2 of its circumscribing rectangular prism. 


ON REVOLOIDS. 87 


PROPOSITION XV. THEOREM. 


If any solid, formed by the rotation of a conic section about its 
axis, that is a spheroid, paraboloid, or hyperboloid. be cut by 
a plane in any position ; the section will be some conic sec- 
tion, and all the parallel sections will be like and similar 


jigures. 


Let ABC be the generating section, or a section of the given 
solid through its axis BD, and perpendicular to the proposed 
section AFC, their common intersection being AC; and GH 
be any other line meeting the generating section in G and H, 
and cutting AC in E ; and erect EF perpendicular to the plane 
ABC, and meeting the proposed plane in F. 


Then, if AC and GH be conceived to be moved continually 
parallel to themselves, will the rectangle AE x EC be to the 
rectangle GE X EH, always in aconstant ratio; but if GH be 
perpendicular to BD, the points G, I’, H will be in the circum- 
ference of a circle whose diameter is GH, so that GE x EH 
will be = EF’; therefore AE xX EC will be to EF’, always 
in a constant ratio; consequently AFC is a conic section, and 
every section parallel to AFC will be of the same kind with it, 
and similar to it. 


Cor. 1. The above constant ratio, in which AK x EC is 
to EF?, is that of KI? to IN?, the squares of the diameters of 
the generating section respectively parallel to AC,GH ; that 
is, the ratio of the square of the diameter parallel to the sec- 
tion, to the square of the revolving axis of the generating 

lane. 
; This will appear by conceiving AC and GH to be moved into 
the positions KL, MN, intersecting in I, the centre of the ge- 
ernating section. 


w3 


88 ON REVOLOIDS. 


Cor. 2. And hence it appears, that the axes AC and 2EF 
of the section, supposing E. now to be the middle of AC, will 
be to each other, as the diameter KL is to the diameter MN 
of the generating section. 


Cor. 3. If the section of the solid be made so as to return 
into itself, it will evidently be an ellipse. Which always 
happens in the spheroid, except when it is perpendicular to the 
axis ; which position is also to be excepted in the other solids, 
the section being always thena circle: inthe paraboloid the sec- 
tion is always an ellipse, excepting when it is parallel to the 
axis ; and in the hyperboloid the section is always an ellipse, 
when its axis makes with the axis of the solid, an angle greater 
than that made by the said axis of the solid and the asymptote 
of the generating hyperbola ; the section being an hyperbola 
in all other cases, but when those angles are equal, then it is 
a parabola. 


Cor. 4. But if the section be parallel to the fixed axis BD, 
it will be of the same kind with, and similar to, the generating 
plane ABC ; that is, the section parallel to the axis, in a sphe- 
roid, is an ellipse similar to the generating ellipse ; in the para- 
boloid, the section is a parabola similar to the generating para- 
bola ; and in an hyperboloid, it is an hyperbola similar to the 
generating hyperbola of the solid. 


Cor. 5. In the spheroid, the section through the axis is the 
greatest of the parallel sections ; but in the hyperboloid, it is 
the smallest ; and in the paraboloid, all the sections parallel to 
the axis, are equal to one another.—For, the axis is the great- 
est parallel chord line in the ellipse, but the least in the oppo- 
site hyperbolas, and all the diameters are equal in a parabola. 


Cor. 6. If the extremities of the diameters KL, MN, be 
joined by the line KN, and AO be drawn parallel to KN, and 
meeting GEH in O, E being the midde of AC, or AE the 
semi-axis, and GH parallelto MN. Then EO will be equal to 
EF, the other semi-axis of the section. 

For, by similar triangles, KI: IN :: AE: EO. 

Or, upon GH as a diameter, describe a circle meeting EQ, 
perpendicular to GH, in Q: and it is evident that EQ will be 
equal to the semi-diameter EF. 


Cor. 7. Draw AP parallel to the axis BD of the solid, and 
meeting the perpendicular GH in P. Then it will be evident 
that, in the spheroid, the semi-axis EF = EO will be greater 
than EP; but inthe hyperboloid, the semi-axis EF = EO, of 
the elliptic section, will be less than EP ; and in the parabo- 


loid, EF’ = EO is always equal to EP. 


i» 


7a 


ON REVOLOIDS. | 89 


Scholium. The analogy of the sections of an hyperboloid te 
those of the cone, are very remarkable, all the three conic 
sections being formed by cutting an hyperboloid in the same 
position as the cone is cut. 

Thus, let an hyperbola and its asymptote be revolved toge- 
ther about the transverse axis, the former describing an hyper- 
boloid, and the latter a cone circumscribing it ; then let them 
be supposed to be both cut by a plane in any position and the 
two sections will be like, similar, and concentric figures ; that 
is, if the plane cut both sides of each, the sections will be con- 
centric, similar ellipses ; if the cutting plane be parallel to the 
asymptote, or to the side of the cone, the sections will be para- 
bolas ; and in all other positions, the sections will be similar 
and concentric hyperbolas. 


That the sections are like figures, appears from the forego- 
ing corollaries. That they are concentric, will be evident 
when we consider that Cc is = Aa, producing AC both ways 
to meet the asymptotes in a@ andec. And that they are similar, 
or have their transverse and conjugate axes proportional to 
each other, will appear thus: Produce GH both ways to meet 
the asymptotes in g and h; and on the diameters GH, gh, 
describe the semi-circles GQH, gRh, meeting EQR, drawn 
perpendicular to GH, in Qand R; EQ and ER being then 
evidently the semi-conjugate axes, and EC, Ec, the semi-trans- 
verse axes of the sections. NowifGH and AC be conceived 
to be moved parallel to themselves, AE X EC or CE?2, will 
be to GE X EH or EQ?, in a constant ratio, or CE to EQ 
will be a constant ratio; and since cE is as Eg, and aE as Ek 
ak X Ec or cE?, will be to gE xX EA or ER?, in a constant 
ratio, or cE to ER will be a constant ratio; but at an infinite dis- 
tance from the vertex, C and c coincide, or EC = Ee, as also 
EG = Eg, consequently EQ is then = ER, and CE to EQ 


a 


7 


90 — ON REVOLOIDS. 


will be = cE to ER ; but as these ratios are constant, if they 
be equal to each other in one place, they must be always so ; 
and consequently CE: Ec: : QE: ER. 

And this analogy of the sections will readily be recognised, 
when we consider that a cone is a species of the hyperboloid ; 
or a triangle a species of the hyperbola, whose axes are infi- 
nitely small. 


PROPOSITION XVI. THEOREM. 


If Sl be the semi-diameter belonging to the double ordinate 
AEC of the generating plane, AKC being the diameter of the 
section AFC, conceived to be moved continually parallel to 
itself ; and x denote any part of the diameter SI, intercepted 
by I. the middle of AC, and any given fixed point taken in 
SI; then will the section AFC be always as a + bx + cxx ; 
a, b, c, being constant quantities ; b in some cases affirma- 
tive, and in others negative ; c being affirmative in the hyper- 
bola, and negative inthe ellipse, and nothing in the parabo- 
la; anda may always be supposed to denote the distance of 
the given fixed point from the vertex s. 


In any conic section, AC? is as a + bx + cx; but all the 
parallel sections are like and similar figures, and similar plane 
figures, are as the squares of their like dimensions ; therefore 
the section AFC is as AC?, that is, asa + br + caz. 


Cor. If the given fixed point, where x begins, coincide with 
the vertex s, then will a be equal to nothing, and the section 
will be as bx + cx, or as x + dra, in the hyperbola and ellipse, 
and as bx, or as 2, in the parabola. 


The subject of hyperbolic revoloids and hyperboloids will 
be considered in another place ; we will here add a few gen- 


eral scholia and formule in relation to cylindric and conical 
ungulas. 


ON REVOLOIDS AND UNGULAS. 91 


af 


SCHOLIA AND FORMULA IN RELATION TG CYLINDRIC AND CONI 
CAL UNGULAS. 


Scholium. It has been shown (Prop. 
VI., Cor. V,) that an ungula pertaining 
toa right revoloid is equal to 2 its cir- 
cumscribing prism, and Prop. III, that its 
curve surface is equal to that of the adja- 
cent side of its circumscribing prism ; 
and because the ungula FHKD is equal 
to an ungula of a right revoloid, if the 
intersection HK of the plane FHK, DHK 
passes through the axis of the cylinder ; 
therefore the convex surface of the ungula 
FHKG is equal to Fh x FD; let GPoD 
be an ungula cut from the former by the 
plane GPo, parallel to the base of the for- 
mer; and the convex surface of the 
ungula GPoD is = the surface FHKD—the convex surface of 
the segment HKFGPo, and this may be divided into two por- 
tions ; viz., the curve surface on the segment PoNn FG, and 
the convex surface of the segment oNn PHK, the latter of 
which is equal (rK X FD + 7K x FD) = 2rKXFD, and the 
former is equal the arc PGo XGF or oN, 

It will be perceived therefore that, although we have the 
quadrature of the whole convex surface of a revoloidal ungula, 
and also of any portion NKo in absolute terms, yet any por- 
tion oGFN, or oGPD is known only in terms of the arc of the 
circle, and consequently depends on the circle’s quadrature, 
but we are enabled by the principles referred to in this scho- 
lium, to determine both the surface and solidity of any portion 
PBF, with the same degree of accuracy as we have the sur- 
face and solidity of the whole cylinder, from which the ungula 
is derived, and since a revoloidal spindle is conceived to be 
made up of partial ungulas, the revoloidal spindle is suscepta- 
ble of the same degree of accuracy in its determination. 

Let FD = h the altitude, and AG = Hh = r the radius of 
the cylinder from which the ungula is derived ; the whole 
convex surface of the ungula is = 2rh - . - - (i) 

Let k = the altitude of the ungula GPoD, and let ot = half 
the chord oP = c, then will the surface ONK = (r—c) X h 
= rh—ch - - : ‘ £ Z c (2) 

Let the arc PGo = p, and the surface oNGF Px, will be = 
Beek) = D0 De eee ee a | Se 


92 ON REVOLOIDS AND UNGULAS. 


Hence the surface of the ungula PGoD = 2rh — rh+ch — 
pht+pk =rh+ch—ph-+ pk - - - - (4) 
Let an ungula ELMD whose base is er than half that 
of the cylinder be considered: if EF = FG, then will the con- 
vex surface of this, equal that of the ungula FKHD + the 
arc (HFK or 3 cr) X EF + ier CS or cP — rh + ch— 
ph + pk - : - (5) 
The solidity of the bi FKHD j is se ere he ainsi (6) 
Let a = the base PGo of the ungula GPoD, and the solidity 
will be (rh + ch ree + pk) i r—a xX 3 (h—h) (Prop. Vs 
B. Il.) = ir?h + rch — lrph + irpk — ah +. 1 ak (7) 
lf A Safes the baile of the ungula ELMD, and eo — Hts 
curve surface, its solidity will be S X S Tee aX (Gh—3h) 
Or enh hae : . i (8) 


Sch. 2. Let AEBF be the base of a 
cone or any other pyramid, right or 
oblique; AV Ba section through the 
vertex by a plane perpendicular to 
the base; EVE, ECF two other 
sections perpendicular to AVB, 
the former through the vertex, and 
the latter through the side at C, 
between V and B. On AB let fall 
the perpendiculars VH, CI; and on 
DC the perpendiculars VK, BL, 
draw CG parallel to AB, meeting 
AV and VH in G and M. 

Then it is evident that EFBV is a pyramid, whose base is 
EFB, and whose altitude is VH ; let the base be called A and 
the altitude a, and the solidity will be 1 Aa; and it is evident 
that EFCV isa pyramid whose base is EFC and altitude VK, 
let the base be called B and the altitude db, and its content 
will bes 1 Bd, it is evident also that the ungula EEFBC = } Aa 
—! b. 

But, by similar triangles, ABV, GVC, it is 


, 1 AB x Cl 
— Cl or HV—VM :: AB: HV, ora erg aint 
my GC x CI 
also AB—GC: CIl:: AB: HV:: GC: VM =. 3- Ge ; 
a DC: BD:: (by the similar triangles ICD, DBL) CI: 
: (because of the similar triangles BCI and CVM, VKC 
GC x CI GC x CI x DB 
iin ‘cBL) JESU ST eT hig Rd DC x (AB—GC) 


ON REVOLOIDS AND UNGULAS. 93 
Therefore the ungula EF BC will be 
Cl GC x DB 
eh OC Xx (A xX AB—Bx—jjq—); - - (1) 


which is a general formula for the ungula of any pyramid. 


If the base be circular, or the pyramid a cone, and the 
angle CDB be less than the angle VAD; or which is the same, 
if CD and VA, produced, intersect in N; the section ECF 
will be a segment of an ellipse, whose transverse axis is CN, 
and conjugate Y” (NO x GC), NO being drawn parallel to 
AB, and meeting VB produced in O. And then the formula 
will become 


1C] GC 
ee x (AB X circular segment EBF — el x 


Re CI 
elliptic segment ECF) =TRBoGC x AB xX circular segment 


GC B (N 
BBR z “ x me Aw) X circular segment, 


whose diameter is CN, and height CD) =, since sim. seg. are 
pint 1CI 
as the squares of their diameters, ABoGé x (AF x circular 
GC x DBx CN x y NO x GC 


segment EBF— DC x AB? ——— xX seg- 

ment of the circle AEBF whose height ee St) = the 

content of the elliptic ungula EFCB - . ° . (2) 
But, by similar triangles, GC—AD : DC::GC:CN = 

GC x CD a) SG OreDB 0 

CrewAD ; and GC—AD :DB::GC : NO =GE_lap Which 


values of NO and NC being substituted in the above expres- 

sion of the elliptic ungula, will give 

—2—__ x [AB circular segment EBF Sea, We eae 

ap—ac% t wchangtd chy ie | AB? *(GC-AD) 

x segment of thecircle AEBF whose height is 

AB x (GC—AD), jh 
Sonne D7 

Ge DB 3 ii 
ae =—— )* x segment of the circle AB whose 


x [D x circular segment EBF— 


p: * ‘Bp=D+¢ 

D x (DB — 
height is ee, - > - 3 (3) 
= the elliptic ungula EFCB; putting A for the height of the 
ungula, D and d for the diameters of the base and end, or top, 
of the frustum, respectively. 


94 ON REVOLOIDS AND UNGULAS, 


| pee 
D—d 
conic frustum, the remainder will express the complemental 


If this value be taken Fro ht ha = the whole 


— 2 —d’) x n — D circular 

3) RB fond BD ; 
segment, whose height isp - + py X (fre oeel 
BD—D+d th 


segment whose height is——— rr a TH a x —nd? + Dx 
3 
segment whose height 1 iy ak bed xX seg- 
BD—D ise 
ment whose height is “ped . ~ 1 cane) 


If the points D and A coincide, the section EFC becomes, a 
whole ellipse, and the formula above, become } Dhn X 
D*—d./Dd 

Dita. wat eee elliptical ungula ACB - - - - - (5) 

And the complemental ungula ACG = dhn X 
D vy Dd—d’ 

TADEST Fe ITAY, ae oc woth tbat Io MetabpecadtD 

If the angle CDB be equal to the angle VAB, the section 
will be a parabola, whose axis is CD, and base EF =2y 
(AD x DB) =2 v(D—d) x d, and its area, by prop. VI, B. I, 
=2DCXEF =4DC vy (D—d) 1) d; and therefore the ex- 


an Dd—ad? 
presssion becomes De Fie (Dxsegment EBF ——, De” 


+ DCO VDd—d@ = the ee ungula EFBC - - - (7) 


If this be taken from 1 hn ee, the remainder will ex- 


press the megs tk ungula x bee i Viz. 


3 


yh X($dVDd—a@ 55 stp 7 X segment whose height 
is D? 

1h 
or HD 1 x(( (D—d) 4./d (D—d) —nd*+ D Xsegment whose 


aD 
height is—7) - - - - -- +--+ +--+ 


* 
5 


ON REVOLOIDS AND UNGULAS. 95 


If the angle CDB exceed the angle VAB, the N 
section will be a an hyperbola, whose transverse 
(a CGx CD 
axis is CN, but the transverse axis = FG pees 
; dxXCD 
CP being drawn parallel to VA, or ppp 
DR DB 


and the conjugate also =GC v PD =d/H-7_DB 


and the area of the hyperbolic section substituted in the gen- 
eral formula (1) will give the solidity of the hyperbolic ungula. 


Scholium 3. If from the perimeter of the section EFC 
there be projected, the surface EIF by y 
perpendiculars AP,CI, &c., to the base of 
the cone, the surface EBFIE on the base, 
will be to that, perpendicular above, viz., 
ECFB on the surface of the cone as OB to 
AV; as the radius of the base to the slant 
height of the cone, (Prop. XII, B. II.,) 
moreover the area of the section EFC is 
to the area EIF as DC: DI; hence the 4 


convex surface FCEB = S = = ms 
TS Ae ee ee a cn : 
And the area FIE Sia FCEF 


DC 
Let ECEF=B and the expression for the surface of the base 
DI 
FIE of the ungula FIEC becomes Da * DI Ean a) 
Letthe base EFB of the ungula EFBC be called A, and 
the area EBFIE will be Mae x B- - - - - - (8) 
CB DI 
And consequently S = TB * A— Da * B - - - (4) 


If from ne x base AFBE of the cone) = the convex sur- 
face of the whole cone, there be taken that of the ungula found 
above ; the remainder ay Xx (PAEF +p, x FCEF) - (5) 
will express the convex surface of the remaining part EFCVA of 


the cone. And if from the value Jast found, there be taken OB 


V 
x circle CG, = aS xX circle GC the convex surface of the 


+ 
5 T4 is 


96 ON REVOLOIDS AND UNGULAS. 


% 


cone GVC, the remainder on x (TAEF x DC x FCEF — 


circle GC) - - - = gir (6) 
will express that ue the SOUIICHiCeE EFAGC. 
If the section FCE be an ellipse, the surface of the ungula 


3 : I 
will be Nea X (circular segment FBE — = X elliptical seg- 


VB GC’ x DI 
ment ee ; and hence OB AB? x (GC—AD) 


wie “Ap * (segment circle AB, whose height is AB X 


x EBF— 


= / = x (eer x 


D2 


DB (D—d) ~ Winn oh 5 Mote ; ; 

DB-(Dowa) * / DBO oii) x seg. cir. AB, whose 
B—D--d 

heightis D x hi. huge ee yi eg eee OL 


And the value of the surface EFCVA, (Formula 5,) will 
VB 
become, OB * (circ. seg. FAE X _ x ellip. seg. FCE) (8) 


.. _ v(4h?+(D—d)’) ad. 2D+d—AD 
Or itis = —~ yg, (PAE + 7 X dA 

oe 7 d—A 
S=B x seg. circle AB, whose height is D x——7— 
ce (DY 


And the value of the surface of the complemental ungula, 
| VB ; 
(Formula 6,) will become 6p B (cir seg. FAK + pa * ellip. 


seg. FCE — circle CG) = v(t +(D~) 


2 1 — D— AD 
& x =m Jun AD * 88: circle AB, whose 


x (-n@+ FAE + 


—AD 


height is D x “72 oo) 4.) Ce a i a ETH) 
When D coincides with A the expression will become, 
VB j _ n/(4h?+(D — d)’) 
D+d 
x (D:——* ya) for the convex surface of the ungula 
RE ie eye = 8m ee ye mole ee era a) 


VB GC 
And Gg X Aly(AB x GO) x n=2n x VBx AIA/ 7p 


2 
, e ; 
ON REVOLOIDS AND UNGULAS. 97 


4h? D—d)’) D+d 
a BS 0! a /Dd for the oblique cone 


AV. os i ws = 
Also a x (AIV(AB x bani 
_ / (4h? + (D—d)?) 

eo. Dr ar 


Xx ee /(Dd—d?)) for 


the complemental elliptic ungula ACG. - - - - - (18) 

If DC be parallel to AV, or the section a parabola; since 
its area Bis 4 DC x DF = 4 DC y(AD x DB,) the general 
formula for the ungula will become 
Ze x seg. (FBE— 4 DI y(AD x DB)) = Ae 
x [seg. FBE to height DB 2 (D — d) y(d(D— @))) for the 
convex surface of the parabolic ungula FEBC. - (14) 

And the expression in Formula 5 will become, 

214.([D—.qd)2 

op X (eg. FAE + $DI v(AD x DB)) = en 
x [(seg. FAE to height AE + $(D—d) v(d(D—d))] for 
that of the part AEFCV. 

Also, that in Formula 6, will be ve x seg. FAE + 4DI 

2 eS 2 

v(AD x DB) — AD: x 2) eee 
ment FAE to height AD + 2 (D—d) yv(d(D—d))’— nd, 
for that of the complimental parabolic ungula FAEDG. (15) 

If the angle CDB be greater than the angle VAB, or the 
section be an hyperbola, its area being found, and substituted 
for B in the general formulz, will give the surfaces ofthe hy- 
perbolic ungulas. 

If the hyperbolic section be perpendicular to the base, DI 
will vanish, and the expressions will become, 
vee ae ae) x seg. of cir. AB, whose height is 3 a 
for the curve surface of the perpendicular ungula CIB. (16) 

4h? +(D—d 
ee x seg. of the circle AB, height 
that of the remaining part FAYE Ge Lic il Ree RS 8 Ty 8 

v (4h? +(D — d) ) 

18? a 
d 

pets — nd*,) for that of the complemental perpendicular un- 


ee ALGO St) ea a ONS! oe (18) 


he 


xX seg- 


, for 


X (seg. of the circle AB, whose height is 


98 ON THE RECTIFICATION AND 


BOOK IV. 


ON THE REVOLOIDAL CURVE, THE RECTIFICATION OF THE 
ELLIPSE, AND OTHER CURVES, AND ON THE QUADRITURE 
OF THE CIRCLE, &C. 


DEFINITIONS. 


1. Tue revoloidal curve is the curve forming the contour of 
one of the facial sides of a revoloid; since this designation may 
apply to any revoloid, therefore, if the revoloidal curve is 
mentioned without reference to the species, the curve of a 
right revoloid is understood. 

2. If the revoloidal surface is extended on a plane, its con- 
tour is called a plane revoloidal curve ; and the surface is called 
a plane revoloidal surface. 7 

3. The vertices of a plane revoloidal 
surface are the two angular extremities, 
as D and E. 

4, The vertical or transverse axis of the 
plane revoloidal surface, is the right line 
drawn through the vertices, as DE. 

5. Its conjugate axis is a line drawn at 
right angles to its transverse, which it bi- 
sects, terminating in the curve, as AB. 

6. A quadrant of a revoloidal surface is 
a portion cut off by the two axes, as ACD 
or BCD. E 

7. Any area bounded partly by curves and partly by right 
lines, is sometimes called a mixtilineal area or space. 


QUADRATURE OF CURVES. 99 


PROPOSITION I. THEOREM. 


The vertical length of one of the plane surfaces of a quadran- 
gular revoloid,is equal to the semi-cireumference of its in- 
scribed cirele, and the length of any double ordinate to its con- 
jugate diameter, is equal to the arc of that circle cut off by 
such ordinate toward the extremity of the conjugate diameter. 


Let ADBE be a plane surface from a 
quadrangular revoloid, and AFBG its in- 
scribed circle, and the vertical length DE of 
the revoloidal surface will be equal to the 
semi-circumference FBG, and the length of 
the double ordinate HI will be equal to 
the arc LBM cut off by such ordinate. 

For since (Def. 8, B. III,) the vertical 
section of a right revoloid through the 
centre of its opposite sides, is a circle, and 
since in a quadrangular revoloid this cir- 
cle is such as may be described on a dia- 
meter equal to the conjugate axis of the revoloidal surface, 
and because each of the facial surfaces of a revoloid extends 
from one vertice to the other, passing through half the circum- 
ference, it follows that its length, DE, is equal to half the 
length of that circumference which is also equal to the semi- 
circumference FBG. 

Again, since any ordinate HI drawn parallel to the trans- 
verse axis DE, is the representative of a parallel to a vertical 
section through the line DE, it follows that the section formed . 
by a plane passing through the ungula of which this face is the 
surface, is a segment of a circle whose chord terminates in the 
curve forming the edge of the ungula, and the section of this 
ungula is similar to the section of its contiguous ungula by a 
plane perpendicular to this section; which section, through 
the contiguous ungula, may be represented by a segment 
LBM ; for a quadrangular revoloid has its sides at right an- 
gles to each other in a plane perpendicular to the transverse 
axis. Hence, if HI is equal to the are containing a segment 
equal to the segment LBM. it is therefore equal to the length 
of the arc LBM. . 


A 


100 ON THE RECTIFICATION AND 
PROPOSITION Il. PROBLEM. 


To make a plane projection of the facial surface of a right 
quadrangular revolovd. 


With a radius EA, equal the radius of the circle forming a 
vertical section of the revoloid, describe a circle ACBD, and 
draw the diameter AB, and from E perpendicular to AB draw 
the lines EF, EG, each equal to one-fourth of the circumfer- 
ence of the circle, ACBD, or equal to one-fourth of the cir- 
cumference of the revoloid. Divide these lines into any num- 


Th 

ber of equal parts, as 1, 2, 3, 4, &c., on the line EF. In like 
manner divide each quadrant of the circumference into the 
same number of equal parts, 1. 2, 3, 4, &c. ; through the divi- 
sions on the circumference draw lines from 1, 2, 3, &ce., parallel 
to DC, and through the divisions on the line EF draw lines both 
‘ways parallel to AB, as 7g, 6f, 5e, &c., and where these lines 
meet the former corresponding lines through the divisions cor- 
responding to the same numbers, will be points in the curve 
forming the boundary of the surface, through which, if a curved 
line a, b, ¢, d, e, f, g, &c. is drawn, this line will represent the 
revoloidal curve, and the space enclosed will represent the 
plane surface of a right quadrangular revoloid. 


Scholium 1. The nature of this curve is such, that, as it pas- 
ses off at the vertices, it reproduces itself again, passing into 


% 


4 


QUADRATURE OF CURVES. 101 


another curve of the same character but of opposite curva- 
ture, for where it passes the vertices F and G, the signs be- 
come changed from positive to negative and from negative to 
positive, so that the curve is reproduced indefinitely as the 
axes EF and EG are continued. 


Scholium 2. The revoloidal curve passes into and becomes 
identical with the circle while passing the extremities of the 
diameter, but its fluxion carries it out of the circle as it leaves 
these points, and it becomes incorporated with and identical 
with a right line as it passes off at the extremities of the axis, 
but its fluction carries it out of the right line as it becomes 
extended. 

For let the diameter AB be produced each way to H and I, 
so as to be equal to FG, and from the extremities of these 
lines draw HF, IF, IG, GH, forming a square circumscribing 
the revoloidal surface; let HF be extended to M, and IF be 
extended to L, then these lines so produced will cross each 
other at right angles in I’, and form an angle with the axis FG 
of 45°. Let the revoloidal curves extend to P and P; now 
if the vertice F be brought to its natural position, on the re- 
voloid, these curve lines evidently cross each other at right 
angles also, and at the point of contact form an angle of 45° 
with the axis, which is the same as that formed by the right 
lines ; hence these curve lines agree with the right lines, HM 
and IL, at that point both in position and inclination, and there- 
fore are identical. 

And also, as the revoloidal curve passes into and occupies 
the space of the circle at the extremities of the diameter, A 
and B having, in its original position, formed a part of the 
circle at that point, and the position of that point not having 
been changed in referenve to the axis or diameter AB, it fol- 
lows, that it is still equal to and identical with the circle at 
those points, but its fluction, or the law of its propagation, 
_ causes it to leave the circle after passing those points, 


Cor. Each of the ordinates through the quadrant, AEF, pa- 
rallel to the axis, EF, is equal to the portions Al, A2, A3, &c., 
of the arc of the quadrant, intercepted by those lines respec- 
tively toward the point A. 


Scholium 3. Hence, this curve is generated by the locus of 
the intersection of two right lines, AC and EF equal the ra- 
dius and semi-circumference of a circle moving uniformly from 
any point IX or A, perpendicular to each other, through 
their respective lengths. 


102 ON THE RECTIFICATION AND 


Let A be the origin, and since CF equal F 
the quadrant AE, hence CF=17, AC=r. 

Let arc 4’ equal any arc Ah, measuring 
the angle ACA, then since Ah=CP it is also E 
é’=y; hence, the equation to the curve is Pp 
y = arc 6/ G 

Let F be the origin, and we have y=QP 
=hG=sin. é’=sin. ACE. 

Then the equation, considering F as the it t 


origin, is y=sin. 4 
PROPOSITION III. THEOREM. 


The contour of a plane revoloidal surface from a right revoloid 
is equivalent to the perimeter of an ellipse, formed by a verti- 
cal section through the angles of the revoloid. 


For the section of a revoloid through the angles is an ellipse 
by definition, and this ellipse terminates the facial surface of 
the revoloid when in its proper position. Now if the cylin- 
dric surface of the revoloid is extended on a plane, its parts 
are not altered in relation to each other ; its vertical length on 
the plane is equal to its length on the cylindric surface of the 
revoloid ; and its conjugate suffers no change, being a right 
line parallel to the axis of the cylindric surface while on the 
revoloid and a right line still when extended ; for the surface 
may be extended in like manner as we would unroll a piece 
of cloth, or a piece of paper, made. to agree with its surface, 
which suffers no contortion of any of its parts in the change, 
but the whole surface is the same in reference to its edges 
after the change as before, and as each facial surface of the 
revoloid extends through half the circumference of the circle 
of the revoloid, viz: from one vertice to the other, each side 
of the surface is terminated by one-half of the ellipse formed 
by a section through the angles, and as the angles of the revo- 
loid cause generally the ellipses to cross each other at the ver- 
tices,. forming a vertical angle also on the facial surface, the 
other side is bounded by one-half of a similar ellipse, so that 
the whole perimeter of the facial surface of a revoloid is 
equal to the perimeter of an ellipse by a plane passing through 
the angles of the revoloid. 


Cor. Hence, the perimeter of a right quadrangular revo- 
loidal surface is equal to that of an ellipse whose conjugate 
or minor axis is equal to that of the transverse axis of the re- 
voloid, and whose major axis is in the ratio to its minor axis 
as the /2: 1. 


QUADRATURE OF CURVES. 103 


For the plane passing through the angles of the revoloid 
forming the ellipse, cuts the plane forming a circular section 
through the centre of the sides at an angle of 45°, therefore, 
the transverse axis of the ellipse is as the diagonal of a square 
of which the vertical axis of the revoloid or its diameter forms 
a side, - 


> PROPOSITION IV. THEOREM. 


If there be described two ellipses concentric with each other, on 
axes of which those of the outer one exceed those of the inner 
one by N, then the two ellipses will not be equidistant through- 
out, but will be nearer to each other at points in the curve be- 
tween the vertices of the axis, than at the vertices. . 


Let AB, ED, and ab, ed, be the 
two axes of two concentric ellip- 
ses, and let the axis ab=Q, and 
ed=R, and let the axis AB=Q 
+N and ED=R+N, then the 
distance between the two ellip- 
ses, between the vertices E and 
A will be less than at those ver- 
tices. 

For, draw the two equal con- 
jugate diameters, HG, FL, and 
also the two Al. fl, 

Then, HG?+FL’, or 2HG’?=AB?+ED? (Prop. XV, of the 
Ellipse,) and 2hI’=ab’*+-ed? 

Now, because the sum of the squares of the axes AB’-++ ED” 
are not greater than HG’?+F'L’, those squares cannot be pro- 
portional (Prop. XVIII, B. I, El. Geom.) hence also. (Prop. 
XXIII. B. L., #1. Geom.) the axes themselves cannot be pro- 
portional. 

Now it is evident, that when the axes AB and ED are nearly 
equal, then also they will very nearly form the extremes of a 
proportion of which the two diameters HG, FL are the means ; 
which is the more nearly true the nearer the two axes are to 
an equality, or the nearer the ellipse approaches to a circle, 
and hence they are more disproportional, the greater the ec- 
centricity of the ellipse. 

Now it is evident, that the inner ellipse, a, e, 6, d, is more 
eccentric than the outer one, AE.BD, since the two axes of the 
inner one are less than those of the outer one by the same 
constant quantity N, (by hypothesis,) hence the conjugate di- 
ameters Al, fl, are more nearly equal to HG and FL than ab 
to AB, or thaned to ED. Hence the elliptical curves are 


104 ON THE RECTIFICATION AND 


nearer each other at any point F or k between the vertices of 
the major and minor axes, than at those vertices. 


Cor. 1. Hence, if about an ellipse there be described a 
curve concentric thereto and equidistant throughout, such curve 
will not itself be an ellipse, but if it is described very near to 
the elliptical curve and equidistant, it may be regarded as an 
ellipse without much error in so considering it, and this will 
be more nearly correct the more nearly the ellipse approaches 
to the circle. 


Cor. 2. If about any curve except the circle, another curve 
be described, in such manner as to be equi-distant in all its 
parts from the former, these two curves cannot be similar 
curves, neither in properties nor in figure, all which may be 
shown by the same reasoning as in the proposition, viz: a 
curve described equidistant from a parabola, either within or 
without, cannot be a parabola, and a curve described equi- 
distant from a hyperbola, cannot itself be a hyperbola ; all of 
which is evident from the properties of those curves, but when 
they are drawn exceedingly near to those curves they may be 
regarded as curves of similar character as those near which 
they are so drawn. 


PROPOSITION V. THEOREM. 


If there be described two curves, one within and the other with- 
out an ellipse, of such kind that they shall both be equi-distant 
from the ellipse, or the ellipse shall be midway between the 
two, then will the space included between the two curves so 
described be equal to the circumference of the ellipse mulii- 
plied by the distance between the two curves. | 


Let FGHK, fghk, be two con- 
centric curves described so as to 
be equi-distant from the elliptical 
circumference AEBD, one with- 
out and the other within, so that 
AEBD shall be midway between 
the two, then will the curvilinear 
space Ff, Gg, Hh, Kk, Ff, be 
equal to the circumference AEBD 
multiplied by their common distance Ff, or Gg. 

For, describe about each of those curves polygons FbmG, 
&c., Anok, &c., fpgg, &c., such that their corresponding sides 
will be parallel, and draw bp, mq, &c., and the surface included 


QUADRATURE OF CURVES. 105 


between the outer and inner polygon will be divided into the 
trapeziums Ffpb, bpqm, &c., each of which is equal to the half 
sum of its parallel sides multiplied by their distance Ff; but ~ 
any side, An of the polygon described about the middle curve 
is equal half the sum of its corresponding parallel sides of the 
trapeziums; hence the trapezium Ffpb is equal to An x Ff, 
and the trapezium bpgm is equal to no X-Ff; and since this is 
true for each of the trapeziums, it follows, that the sum of all 
is equal to the sum of all the sides of the polygon described 
about the ellipse AEBD, multiplied by the common distance 
Ff. And this would be manifestly true, whatever be the 
number of the sides of the polygon described about the ellipse, 
but when the number of the sides of the polygon is indefi- 
nitely increased, the polygon becomes a curve similar to that 
about which it is described. (Prop. XII, Cor. 4, B. V, Ei. 
Geom.) Hence as in the proposition. 


Cor. It is evident, also, that if there be described curves 
within and without a parabolic or any other curve, so as to 
be equi-distant from it, then the space included between the 
outer and inner one will be equal to the curve situated midway 
between them, multiplied by the distance between the outer 
and inner curves, and the same may be aflirmed of the revo- 
loidal curves so drawn. 


Scholium 1. The polygon described about two eccentric 
curves drawn so as to be equidistant throughout, are not simi- 
lar polygons, since the figures about which they are described 
are not similar. (Prop. [V., Cor. 1.) 

Scholium 2. The last two propositions suggest a method of 
rectifying the elliptical circumference, and also of finding the 
lengths of other curves whose quadratures are correctly or 
approximately known. 

For, if about an ellipse AEBD, whose major axis AB=P, 
and whose minor axis ED=Q, another ellipse FGHK be des- 
cribed, whose major axis FH=P-+N, and whose minor axis 
GH=Q+N, and if another concentric ellipse be also des- 
cribed within the former, whose major axis fh=P— N and 
whose minor axis gk=Q — N, we shall have an elliptical rng 
Ff, Gg, Hh, Kk, whose area may be found by subtracting the 
inner ellipse from the outer one, then if this area is divided by 
the distance of the inner and outer circumference the quotient 
will be the elliptical circumference AEBD. 


106 ON THE RECTIFICATION AND 


PROPOSITION VI. PROBLEM. 


Let it be required to find the distance Mm, between two concen- 
tric ellipses, the difference of whose major and minor axes 
are each equal to a quantity N. 


Let HGPK, bghk be two 
concentric ellipses, the dif- 
ference of whose axes, HP 
—bh, is N, also GK—gk 
=N; draw the equal con- 
jugate diameters ML, NI, 
ml, ni; draw the right co- 
ordinates MN, mn, draw 
mt parallel to HP, and join 
Mm, which will be the distance between the ellipses at the 
points M, m; we shall have sC?=13PC’ and sM’=}CG’, also © 
EC?=1AC? and st??=3Cg"* (Prop. XXI. Cor. 1 of Ellipse.) and 
sC—EC=sE or mt, and sM—st or mE=Mt. Hence VM??4+-mé* 
=Mm=the distance of the two curves at the points M, m. 

Let the axis HP=25, and GK=19, and if N=, then 

bh—=23', and gk=—1F 
and sC=8.83175 EC=8.13172, 
hence im=.70003 

sM=6.71751 and mE=6.00999, 
hence Mi=.70752 
and Mm=.995302. 


Scholium. 1. But because the line Mm is not perpendicular to 
the curve AFRD at the point of contact, but very nearly per- 
pendicular to MW,, it is therefore greater than the true distance 
of the curves, which we will sappose is wv, to find which, we have 

; Se 

YC EG? and WC= C 

YE=YC—CE. and Ws=WC —Cs 

mY =Vmb?-+EY*, MW= Vv Ms?+ Ws? 
Therefore, we have the sides of the right-angled triangles mEY, 
MsW given to find the angles W and Y, the difference of 
which, when found, is equal to the angle made by the lines MW 
and mY with each other produced. Then we have a right- 
angled triangle Mm, MW produced. and mY produced right- 
angled at M, to find the sides MW produced, and mY pro- 
duced, and also the base wv of an isusceles triangle having the 
same vertical angle, which line uv will be the shortest dis- 
tance required. 


QUADRATURE OF CURVES. 107 


Scholium 2. Let the equal conjugate diameters IN, LM, in, 
im, be drawn, and the difference of the semi-diameters C1—Ci 
will be very nearly equal to the distance It of the curves at 


the points 1,7. But IN= ViPH?+!GK%, and in= Vi1bA+2¢k', 
hence, ViPH?+:GK?— V1 6h*+1¢h?=the distance Ii very 
nearly. 

Let the axis AR=24, and FD=18; then, if we make PH 
=25, and GRh=19, IN will==22.2036. 
Also, we may have bh=23, gk=17; hence, in or ml=20.2287. 


22.2036 — 20.2237 ; 
Therefore, ———-———-———— =.9899 =the difference CI and 


Ci, which is nearly equal the true distance of the curves 
through the point A, which will be more accurate as the axes 
appruach equality, and will be approximately true till the ec- 
centricity of the curves becomes very great. 


PROPOSITION VII. PROBLEM. 


To find the length of the ell ptical circumference, approximately. 


It has been observed, (Prop. V. Scholium,) that the cir- 
cumference of an ellipse inscribed between two other concen- 
tric ellipses, is equal to the area or space included between the 
two divided by their distance from each other ; but since the 
distance of the curves is not constant in every part, we must 
take their average distance 

If Gg. PA, (see diagram to Prop. VI.) be the distance of the 
two extreme ellipses through the lines of their axes, and uv the 
distance through the point e, then the average distance will be 
very nearly equal }Gg+ uv; hence, if the area of the ring within 
the exterior and interior circumferences is divided by }Gg-+1upv, 
the quotient will be the length of the whole circumference 
AFRD ; or if a quadrant of the ring is divided in like man- 
ner, the quotent will be the length of a quadrant FeR of the 
circumference. 

If we assume the axes AR=24, FD=18, as in the last pro- 
position, and the axes of the other ellipses as there assumed, 
we shall have for the area of the greater ellipses, PH 
X3GH X7r=12.5X 9.5 x+=373.06381; and the area of the 
smaller ellipse $2kX 1bhX i=11.5X8 5X* =507.09042, there- 
fore the area of the ring is equal to 65.97339. 

( : 

Let this be divided by et at the average distance as 
found in the last proposition, and we have 66.128 for the cir- 
cumference of the ellipse AFRD. 

Let the area be divided by the distance as found in (Sch, 


108 ON THE RECTIFICATION AND 


9899+ 1 
2. Prop. V,) and we have65.97339~+ aa 


the elliptical circumferance, which is true to four places of 
figures, and very nearly to six; hence, this mode of compu- 
ting the elliptical circumference is sufficiently accurate for any 
ordinary calculation. 


= 66.3115 = 


Scholium. The length of the parabolic or hyperbolic arc 
may be approximately determined in the same manner as is 
here suggested for the ellipse. 


PROPOSITION VIII, PROBLEM. 
Let it be required to find the length of a revoloidal curve. 


Let ADBE be a revoloidal curve 
from a right quadrangular revoloid 
whose length is required; let two 
other concentric curves KLMR, 
FGHI, be described on each side 
of the first and equidistant there- 
from, and if these two last des- 
cribed curves are at a small dis- 
tance only from the former, they 
will be very nearly revoloidal 
curves likewise. 

Let aa, the axis of the revoloid 
from which the surface ADBE is 
supposed to be taken, equal P; 
and since, by hypothesis, ADBE is 
the surface of a right quadrangular 
revoloid, the conjugate is also equal 
to P, and the vertical length DE of the revoloidal surface will 
be equal }+XP ; since DE is equal to half the circumference 
AaBa of the revoloid. Now, let N be the distance AK, or AF, 
that the concentric curves are proposed to be drawn; and 
since the angles DLn and LDn are each=45° in a right 
quadrangular revoloid, make the semi-circumference of the 
revoloid from which the surface KLMR is taken, equal 
daP+2V Dn? + Ln? =30P+2V2N?=!n', and make the circum- 
ference of the revoloid from which the surface FGHI is sup- 
posed to be taken, =}+P —2VDn?+ Ln?=!0¢P—2V QN = 14, 
then may P’=the diameter of the greater revoloid, and P” 
equal the diameter of the lesser, and P+2N equal the conjugate 
axis KM of the larger surface, and P—2N equal the conju- 
gate axis of the smaller. ms 


QUADRATURE OF THE CIRCLE. 109 


And we have (Prop. III. B. IIL) the area of the surface 
KLMR=(P+2N)XP’, and the area FGHI=(P —2N)xP”, 
and (PP’+2P’N) —PP”"+2P"N equal the space KF, LG, 
MH, RI between the inner and outer curve. Let this area be 
divided by the distance KF or N, and the quotient will be the 
length of the revoloidal curve ADBE very nearly. 


Scholium 1. Since the major and minor axes of the outer 
and inner curves vary in very nearly the same ratio, it follows 
that the outer and inner curves must be very nearly similar to 
the central one where the distance KF is small, even if the 
surface possesses a considerable degree of eccentricity, pro- 
tracted in the direction KM ; but the greater the eccentricity 
when protracted in the direction LR, the greater is the simila- 
rity of the concentric figures, and the greater is the accuracy 
with which we can rectify the curve. 


2. Since the perimeter of a revoloidal curve is also the pe- 
rimeter of an ellipse, such as is formed by a vertical section 
through the axis and the angles of a revoloid, this mode of rec- 
tifying the revoloidal curve furnishes also a mode of rectifying 
the elliptical circumference ; and reciprocally. 


PROPOSITION IX. PROBLEM. 


To find the vertical length of the revoloidal surface, and 
consequently the circumference of the circle. 


Let ABC be the quadrant of 4 
one of the facial surfaces of a 
quadrangular revoloid, and BCD 
a quadrant of the inscribed cir- y 
cle; AC the semi-transverse, M2 
and CB the semi-conjugate axis ; 
extend CB to FI’, making it equal 
to AC, draw AF, and the trian- 
angle ACF will be equal to one- 
fourth of the square circumscrib- 
ing the whole revoloidal sur- 
face. From the extremities of © 
the radii CD and CB, draw the gB F 
lines DE and BE, perpendicular to those radii respectively, 
then will BCD be a square circumscribing the quadrant equal 
to one-fourth of the square circumscribing the circle. 

On the line BE take any distance Ba, and from the point a, 
draw the line ab parallel to the semi-conjugate, BC ; then set 
off from C on the transverse CA, the distance Cc equal the are 


110 ON THE RECTIFICATION AND 


Bi cut off by the line ab; and draw cd also parallel to BC, 
cutting the curve AB, and from d draw dg parallel to AC, 
then will the rectangle CBab equal the portion of the revoloidal 
surface CedB. (Prop. III-Cor.1, B.III,) and cd=ib=cosine of the 
arc Bi, and Ch=ig=sine of the arc. The portion of the revo- 
Joidal surface CBdc may consist of the rectangle Ccdg, and the 
segment Bdg ; the former of which is equal to the product of the 
line Cc or dg, into the cosine ib or cd; and the latter may be 
divided into the two portions Big, a segment of the circle, and 
the trilinear space Bid, the space included between the circle 
and the revoloidal curve; the former of which is equal to 
the difference between half the product of the arc Bz, or its 
equivalent Cc, into the radius CB; and half the rectangle of 
the sine ig, into the cosine 7b; and the latter may be approxi- 
mately estimated by considering, that the distance between the 
circle and the curve at any point, and in direction parallel to 
the base line id, is equal to the difference between the arc of 
the circle included between such point, and the radius CB, and 
its sine. Thus, the distance id=arc Bi or dg, its equivalent, 
— gi, its sine. Hence, it will be perceived that its value con- 
verges rapidly as we approach the semi-transverse AC. 
Though the ratio of this space is constantly changing with 
regard to its linear dimensions, as the vertical length of a 
conjugate section is varied; yet, when the distance Ba is taken 
very small, its area may be regarded as equal to one-half the 
product of the base 7d, into its vertical height perpendicular to 
that base, viz., into gb. 
Let x=the arc Bi=Cc, 
and s=sine of the arc Bi, 
and c=cosine. 
Then will ce=the rectangle Ccdg, 
and jrz—%3cs=the segment Big, 
fire) re en fos 
Hi we ae eee 
and... .7s=the rectangle BCba. 
Hence, cx+tra— hes+3ra—'rs—icxr+ics=rs. 
By transposing and condensing icx+ra=2rs 
a. are 3s 
Therefore, he PART OEE Ta (1) 
Or if we make the curvilinear space idB=aiB, making the 
mixtilineal space Ccdb equal to the arithmetical mean between 
the mixtilineal space diBCc and diaBCc, we shall have 
xc+ kra—tsc=the space diBCc 
and xc + rs —cs =the space diaBCe 
half the sum of which make=the rectangle 
| CbaB=rs. 


the space Bid, 


QUADRATURE OF THE CIRCLE. 11 
re a Qex+-3 eee eat =rs 
and Qex+tre—{cst+rs. 
3CS+Ts 
Hence, So a a 2.) 


Again, draw fG parallel to 4 
CBor DE, cutting offany arc DL, 
and set off from A the dental 
AH, equal the length of the arc y 
DL. and from the point H, draw b 
the line Hé, perpendicular to the 
axis AC, to cut the curve in v, 
and the trilinear space AHv will 
be equal to the rectilineal DEG, 
(Prop. II, and Cors. B. HI.) 

Now, if we compute the area = 
of this trilinear space AHv in c 2B F 
terms of its sides, those sides may be rectified by its known 
quadrature. The length of the side AH equal the are DL, and 
the area AHv is less than half the rectangle of AH into Hv, 
and greater than }vH*; but will be very nearly un arithme- 
tical mean between the two. 

Let AH Sr=are DL, 

Hv=s=sine of z, 
DG=v=versed sine of x. 
te oes 


Then 9 area AHbp, 
and vr=area DGfE; 
but it has been shown aks the area AHv=area DGfE. 
Lites SL 
Hence, ur=* 5 
Or, 4vr=2z'+2s. 


2 


Ce Ss 
Completing the square vr sta = 4dr 


‘a Nhs 
Hence, . Ce X/ oo ae - - (3.) 


The smaller the arc is taken, the greater the accuracy ; for 
when the arc is taken very small, the sine and arc are very 
nearly identical; when no appreciable error would occur in 
so considering them, and the area AHv would still be found 
between half the square of AH, and half the rectangle of AH 
into Hi; and if the difference of x and s becomes =o, then 

2 


e 
the expression for the surface becomes barely a 


’ 
- ¢? 


’ - 


112 ON THE RECTIFICATION AND 


and pa V re) =! eo ee eS (4,) 
Let the two arcs, DL, iB, be taken such that the diagonal 
At is nearly equal the arc 2B, and if we make AH¢ equal the 
rectangle DGfE, and nBCe equal the rectangle aBCe, the re- 
sult will be a close determination of the length of the arc of 
the circumference ; for the area AH, in such case, is very 
nearly as much in excess, above the area AH», as the area 
nBCc is in defect of the area dBCc, and if these arcs are 
taken very small, any error may be rendered evanescent. 
For this purpose, let z=the arc iB 


and z'=the arc DBL. 
Then let cr—ics+-j}rz=rs 
: J les-ttrs 
=- — -« wa MS OST OR: 
which reduced gives z cLir (5.) 
And obit a! 
Then Se OUT te eee we ow MO 


Whence, if we take an arithmetical mean between the re- 
sults of these two equations, we shall arrive at a very approx- 
mate determination of the circumference, if the arc is taken 
very small. 

For an example, let us take an arc= z,;1,; part of the cir- 
cumference. We have, by trigonometry, the sine of that arc 
=.00025566346=s, and its cosine=.99999996732=c. Hence, 
by formula 1, we have 

ars 
— r-+ic 
Therefore. 00025566346 X 3= 3rs=.00038349819 
and 1+9999996732+2=r+ 7c=1.4999998366 


: | 
And nae .0003834981 9 1.4999998366=.00025566346278 
2 

hence x=.00025566346.278, which multiplied by 24576, gives 
the whole circumference=628318526133 when the diameter 
is=2, or 3.1415926306=7, when the diameter is 1, which 
result is true to eight places of figures, but the 9th should be 5 
instead of 3. 

If instead of this arc, we take that of ;515; of the circumfer- 
ence, the sine of which is=.000157079632033, and the cosine 
=.999999987462994 

3rs=,.000235619448050 
r-+ic=1.499999993831497 


3 


3rg 
2°* _-—000157079632676 
r+ ze 

000 


40 
whence 7=.000157079632676 x ae 3.1415926535.2, 


| 
* . 
yo , 


QUADRATURE OF THE CIRCLE. 113 


which is true to eleven places, but the last figure should be 8 
instead of 2. 

Hence, it will be perceived, that, by taking a very small arc, 
we can rectify the circumference to any degree of exactness. 


Cor. 1. Since in formula (6) }2’=vr; if r=1, then v=}2? (1.) 

Hence, the versed sine may be taken as half*the*square of 
the arc, which is approximately true wheng/the arc, is. small, 
and if very small, may be taken as an accurate determination, 
but as the arc is increased, if we proceed as in formula (3,) 
where rv=12z*+1sz, we have the value of'v or the versed sine, 
equal to | of the square of the arc+1 the rectangle of the arc 
and sine; which is approximately true for any arc of the quad- 
rant, and may be taken for an accurate determination when 
the arc is small. 

Hence this general formula v=12°+1s8e - -- - (2) 


2 
gis 


r+ic 
deduce expressions for the sines, of small arcs, in terms of the 
cosine and arc; and also for the cosine in terms of the sine and 
arc. For from this equation we obtain 


Cor. 2. Hence, also, from formula (1,) z= we may 


s=2z-+ler - - + +. + + - + (1) 
3 
and c=——2 = weaerat) Payee ne Wad eT eey) 


Scholium. 1. If s is taken equal to the sine of 30°, we shall 
avoid in some measure the inconvenience of using imperfect 
decimal terms; for the sine of 30° is equal to half the radius, 
and we then have only one decimal term entering into the ex- 
pression, viz., the cosine which is, 866025, &c.; but we have 
a more difficult determination of the area of that portion of the 
revoloidal curve existing without the circle, viz., the space Bid; 
which, if determined, would lead to the true determination of 
the arc of the circumference. 


2. If the ratio between the quadrant of the circle and the 
portion of the revoloid without the quadrant, viz., if the portion 
ADB is determined, then the ratio of the circumference may 
also be determined. 

Thus, if }r*=the area of the quadrant CDB, and z=the area 
of the space included between the quadrant and the curve, 
viz., ABD, and if the ratio between these terms are known, 
that is, if pee Wet ee 

then trx+z=r (Prop. Ill, B. ML.) 
by uniting extremes and means, 


114 ON THE RECTIFICATION AND 


irnx=2m 
z= — tra 
By substituting this value of z in the former equation irnz 
=mr* — rmx; by dividing and transposing }nv-+}mz=mr ; 
pee Oe 
AT et / Se 


hence of the circumference. 


3. Or if the ratio between the quadrant of the revoloidal 
surface and its circumscribing parallelogram is determined, 
the circumference may be computed also. 

For the revoloidal quadrant being equal to the square of the 
radius, (Prop. III. B. III,) it is in the same ratio to its circum- 
scribing parallelogram, as the radius to one-fourth of the cir- 
cumference, seeing the parallelogram is equal to the product 
of the radius by one-fourth of the circumference. 


PROPOSITION X. THEOREM. 


If a quadrant ABC of a plane revoloidal surface from a right 
quadrangular revoloid be described, and a rectangle ACBD 
circumscribe the quadrant, then if the axis AC be divided 
into any number of equal parts, and if ordinates be drawn 
from the points of division across the rectangle, then the 
parts of those ordinates included between the axis AC, and 
the revoloidal curve BA, will be equal to the sum of a series 
of sines of arcs of the quadrant EB, of a circle in arithmeti- 
cal progression, equal in number to the number of the ordi- 
nates, and the parts of those ordinates intercepted by the curve 
AB, and line CD ; will be equal to the sum of a similar series 
of versed sines, and the sum of the whole ordinates will be 
equal to an equal series of radii of the quadrant EB. 


Draw any ordinate as KG, and from m, 
where it cuts the curve, draw mM paral- 
lel to AC, and let it cut the circumference 
of the circle described on the ordinate 
CB, in x draw nh parallel to mK, and the 
line mK=the line nh, is the sine of the 
arc nl, and because (Prop. II,) the line 
AC=the quadrant EB of the circumfer- 
ence, and the line mM=KC is also=the 
arc nB, the line KA equal the arc nE; 
take Kd=KA, and from d, draw the ordi- 
nate dH parallel to the former ordinate, 
and from the point 27, where it cuts the curve, draw iL, and 


QUADRATURE OF THE CIRCLE. 115 


from the point a where the latter line cuts the quadrant, draw 
ab, which will be the sine of the arc aE=to the part id of the 
ordinate dH ; and because iL is equal to aB (Prop. 1.) and mM 
=nb, and AC=EB, and because AC —mM=mM —iL, En 
=na; and since the same may be shown in reference to any 
other ordinate, drawn from a point on the axis AC, whose dis- 
tance from d toward C is equal to Kd or AK, it follows that 
the parts of equidistant ordinates drawn across the rectangle 
intercepted between the curves AB, and the axis AC, are equal 
to the sum of a similar series of sines of arcs of the quadrant 
EB taken in arithmetical progression; which is the first branch 
of the proposition. 

And since av=1H, is the versed sine of the arc aB corres- 
ponding to the sine aL with its complement ab, and nw=mG 
versed sine of the arc 7B, its complement being mK; and since 
they have been shown to be at distances from each other pro- 
portional to the arcs En, na, &c. ; and since the same may be 
shown in reference to the portion of any ordinate intercepted 
by the curve AB, and line BD, wherever drawn, it follows that 
the sum of the portions of the equidistant ordinates intercepted 
by the curve AB and line CD, is equal to the sum of a similar 
series of versed sines of arcs taken in arithmetical progression ; 
which is the second branch of the proposition. And since the 
lines CA and BD are parallel by hypothesis, the ordinates are 
all equal in length, and equal to the radius CB, hence the whole 
series of ordinates is a similar series of radii. 


Cor. 1. From the preceding demonstration, it appears that 
any line or ordinate drawn from any point on the axis AC, 
parallel to the conjugate CB, and terminating in the curve AB, 
is equal to the sine of the arc on the quadrant BE, cut off by a 
line drawn from the point of termination of ‘such ordinate in 
the curve parallel to the axis AC. 


Cor. 2. Hence, also, the sum of the series of sines is to the 
sum of a similar series of radii as the square ECBF, described 
on the radius, to the rectangle CB, AC, of the radius and are 
of the quadrant. Since the square ECBF is equal (Prop.{lil. 
B. III.) to the surface ABC. 


“ee 


116 ON THE RECTIFICATION AND 
PROPOSITION XI. THEOREM. 


The area of a plane revoloidal surface is to that of its circum- 
scribed parallelogram, as the sum of an indefinite sertes of 
sines in the circle, to the sum of an equal series of radit. 

‘ Let ACBD be a parallelogram circum- 
scribing the quadrant of the revoloidal 
surface ABC, and let EBC be a quadrant 
of the circle, then will the area of the 
quadrant of the revoloidal surface ABC 
be to that of a parallelogram ACBD, as 
the sum of an indefinite series of sines of 
the quadrant BCH, to the sum of an equal 
series of radii. 

For, since all ordinates vb, dH, &c., 
cut the surfaces of those figures in rela- 
tion of their magnitudes in the sections C M iit vcs 
through which such ordinates pass; and since (Prop. XI,) if 
ordinates be drawn through those figures equidistant from 
each other, the portions of the ordinates intercepted by the 
curve and axis, are equal to the sum of a series of sines of ares 
in arithmetical progression for the whole quadrant equal in 
number to the number of the ordinates, and if these ordinates 
are equidistant from each other, the sum of the portions passing 
through either surface, drawn into their common distance, may 
be taken for the surface ; and since the distance of the ordinates 
is equal by hypothesis, both for the parallelogram and revo- 
loidal surface, the portions of the ordinates intercepted by each, 
will be in relation to their surfaces respectively, when their num- 
ber is indefinitely increased, and their distance becomes indefi- 
nitely small. Hence, as the sum of a series of sines of arcs of 
the whole quadrant taken in arithmetical progression, is to the 
area of a quadrant ACB of the revoloidal surface, so is the 
sum of an equal series of radii to the area of the parallelogram 
ACBD; and what has been shown for one quadrant of the 
revoloidal surface, is also true for the whole. 


Cor. Hence, the space BDA without the revoloidal surface, 
is to the revoloidal surface, as a sum of an indefinite series of 
arcs in arithmetical progression to the sum of a similar series 
of sines. 


QUADRATURE OF THE CIRCLE. 117 


PROPOSITION XII. THEOREM. 


If the quadrant AB of a revoloidal curve. be made to revolve 
about its axis AC, and if a plune hemisphere of a quadran- 
gular revolovd be described about the solid so generated, hav- 
ing the same axis AC, then the revoloid will be to its circum- 
scribing prism, as the sum of the squares of a series of sines 

_ of the quadrant, to the sum of the squares of an equal series 
of radii. 


Let an indefinite number 
of planes be passed through 
the revoloid perpendicular 
to the axis, and at equal dis- 
tances from each other, and 
the sections made by these 
planes will all be squares, 
(Prop. 1, Cor. 4 B. III.) and 
their sides will all be equal to 
the ordinates Hm, hn, &c., 
drawn through the intersec-  F C B 
tion of such planes with the vertical sections; and hence the 
side of each parallel section, is equal to twice the sine nh of 
the quadrant corresponding to such section, being = nn. 
Now let each of those parallel planes be extended to HG pass- 
ing through the prism, and it is evident that each of the sec- 
tions of the prism will be squares, whose sides are severally 
equal to twice the radius CB, of an inscribed circle. Now 
the magnitudes of these solids through each section, are evi- 
dently in the relation of the magnitudes of such sections ; and 
if the number of these equidistant planes are indefinitely in- 
creased, and they are indefinitely near together, their sum will 
represent the whole of each of the solids in the relation of 
their whole magnitudes, and since each conjugate section of the 
revoloid is the square described on double the sine, answer- 
ing to the ordinate in reference to the quadrant CEB, and each 

section of the prism is the square described on the line HG, 
equal twice the radius, it follows that the solidity of the revo- 
loid, is to that of the circumscribing prism, as the sum of the 
squares of a series of the sines of the quadrant, to the sum of 
the squares of an equal co of radii. 


Cor. Hence, the solidity of the space between the surfaces 
of the revoloid described as above, and that of its circum- 


118 ON THE RECTIFICATION AND 


scribing prism, is to that of the revoloid, as the sum of the 
squares of a series of versed sines of the quadrant, to the sum 
of the squares of an equal series of sines, and this space is to 
that of the prism, as the sum of the squares of the versed 
sines, to the sum of the squares of an equal series of radii. 


Cor. 2. Let the prism and also the revoloids, be divided 
into four quadrants by planes through the vertical axis, and 
passing through the centres of the opposite sides; and the so- 
lid so described. will be truly represented by the value of the 
conjugate parallel sections, passing through them, viz: the 
seginent of the revoloid, will be represented by the sum of the 
squares of the sines; the space between the revoloid and sur- 
face of the prism, by the sum of the squares of an equal series 
of versed sines, and the prism by the sum of the squares of an 
equal series of radii. 


Cor. 3. Since the revoloid has the same ratio to its circum- 
‘scribing prism, as the solid of revolution about which it is des- 
cribed, has to its circumscribing cylinder; the solid formed 
by the revolution of the revoloidal quadrant AB, will be to its 
circumscribing cylinder, as the sum of the squares of a series 
of sines of the quadrant, to the sum of the squares of an equal 
series of radii. ; 


PROPOSITION XIII. PROBLEM. 


Let it be required to find the circumference of the circle from 
the ratio of the sum of the series of sines for every minute 
of the quadrant to the sum of an equal series of radit. 


The number of the series of sines to every minute is 5400 
= the number of minutes in the quadrant, which is the num- 
ber of radii to be compared with the series of sines, and if the 
radius = 1, then 5400 is the sum of the series of radii; and 
the sum of the series of sines to every minute is by Trigono- 
metry = 3438.2467465. 

And (Prop. XI,) the area of the revoloidal surface is to 
that of its circumscribing parallelogram, as the sum of an in- 
definite series of sines to the sum of an equal series of radii; 
but the series of sines to every minute being a definite number, 
and such that the surfaces between the lines may be render- 
ed appreciable, they do not represent those spaces or tra- 


i 


QUADRATURE OF THE CIRCLE. | 119 


peziums in their exact ratios, but represent the longest sides of 
those trapeziums, making up the revoloidal surface, but in or- 
der that they may be true indices of those trapeziums, they 
should be such as pass through the centres, when each would 
be reduced, by a quantity equal to half the difference between 
itself and the next greater one, and as the sum of all their 
differences, is evidently equal to the radius, halt of the sum of 
their several differences is equal to half the radius ; therefore, 
the sum of the natural sines must be reduced by that quantity, 
viz: 3488.2467165 — ,5=3437.7467465, when if we make 
r = radius 
and x= 1 of the circumference, we shall 
have, (Prop. IX., Sch. 3.) 
r:2:: 3437.7467465 : 5400 

Hence, x=1.570.796337= 1 the circumference when 
the diameter is 2, which is true to 8 places of figures, viz: to 
1,5707963, but the 9th figure should be 2 instead of 3. 


Cor. 1. Because the cosine of 60° is equal to half the ra- 
dius, and because the surface of any conjugate section of the 
revoloid is equal to the radius multiplied by the cosine corres- 
ponding to each section, (Prop. XI, \ or. 1.) the sum of the 
sines for 60° is equal to half the sum for 90° or the whole 


quadrant; and consequently, is equal to the sum of the series, 
for the arc from 60° to 90°. 


Cor. 2. Hence, by proceeding as in the proposition, using 
the sum of the sines for an arc of 60° in comparison with a 
corresponding portion of the circumscribing parallelogram, 
the ratio of the circumference of the circle to its radius, may 
be determined by this arc, in the same manner as by the 
whole quadrant. 


Scholium. It will be perceived that the sum of all the natu- 
ral sines of the quadrant, to any number denoting the series, 
may be calculated by reversing the operation, viz: w:r:: 
mr: sum of series of sines minus tr, when nm = the number 
denoting the series: and this may evidently be effected for the 
whole quadrant or any portion of it. 


ie 


” 


120 ON THE RECTIFICATION AND 
PROPOSITION XIV. THEOREM. 


If there be any number of equidistant ordinates of different 
lengths drawn from aright line AB, and terminated by the 
vertices of a polygon, then the area comprehended between 
the greatest and least ordinate, and the right line AB and 
polygonal line CDEF is equal to the sum of all the middle 
ordinates + half the sum of the extreme ordinates drawn into 
the common distance AG. 


For the quadrilateral ACDG 


is = 1 (AC+GD) AG, Bs 
the quadrilateral EDGH = 
1 (GD+HE) AG or GH, and the quadri- 
lateral HEFB =! (HE+BF) HB; hencec 
by addition we have 
(2 AC+ GD +EH+; BF) AG. 
A H B 


PROPOSITION XV. THEOREM. 


If a right line AK be divided into any even number of equal 
parts AC, CE, EG, gc. ; and at the points of division their 
be erected perpendicular ordinates AB,CD, EP, &c., termi- 
nated by any curve BDPS ; and if a be put for the sum of 
the first and last ordinate AB, SK, b for the sum of the even 
ordinates CD, GH, LM, FQ ; and c for the sum of all the 
rest, EP, IR and NO ; then (a+4b+2c) x 1 of the common 
distance AC will be the area, ABSK very nearly 


Through the first three points Q@ Ss 
BDP, let a parabola be con- Fee | 
ceived to be drawn, having its ae 
axis parallel to the ordinates ; 3 
the parabolic area ABPE, (Prop. iy 


VI. Schol .B. 1) will be (AB+ x 
4CD+EP) x 4+ AE = (AB+ 
4CD+EP) x 4+ AC; and when 


z P ¢ aes ie yal ee Ge 
the points of BDP are at no CO GL Tan ue 


great distance from each other 
the parabolic curve will very nearly cioncide with any other 
regular curve, drawn through the same points. 

Let us now take the ordinates EP, GH, IR ; then will (EP 
+4GH-+IR) xX } EG=the area EPRI; and (IR+ 4LM+NO) 


oD 


i‘ | 
QUADRATURE OF THE CIRCLE, 121 


x 1 [L=area IRON. Also (NO+4FQ+KS) x1 NF=area 
NOSK, whence by addition, we have [(AB+KS) + (4DC+4 
Ree tot AE cae Ag BNO Doe ig (21 225-20) 
xt : 


Cor. This theorem may be applied for computing the con- 
tents of solids, by using parallel sections instead of the ordi- 
nate as will appear in Prop IV., Corellaries and Scho- 
ium, B. I. i , 


Schol. It is evident that the greater the number of ordinates 
and the nearer the points DBP, &c., are to each other, the 
more nearly will any curve, drawn through them, agree with 
the parabola, and hence the greater accuracy will be ob- 
tained ; the same remark will also apply to solids. 


Cor. 2. Hence if the area of any space AB, KS is known, 
and the ordinates AB, IR,KS, the value of the line AK may be 
determined. 

For if AB + KS = aand IK = 3, we have, by considering 
the curve as a parabola, whose axis is parallel to KS, the area 
A =(a+ 4b) x + AK, let AK = p, hence we have 

A 1M, 
ln — -—_# ee 2 Bs u 
th tec ae) +, dict ahodetad 6: (1) 

Or, if the number of ordinates is increased, we have by 
the proposition, A= (a + 4b + 2c) i p+ n, w being the 
number of divisions in the line AK, or the number of ordinates 


3nA 
less 1, and we have p = eee - - ” (2) 


PROPOSITION XVI. PROBLEM. 


To find eleven equidistant ordinates to hyperbola between the 
asymptotes, and by means of those ordinates to find the 
area. me 


Taking the equation a?=zy, and assuming a = 10, and the 
first value of z, or the distance from the centre to the first 
ordinates = 10, and if the lower distance is 1 = d, we shall 
have for the ordinates: 

10 10 10 10 10 10 10 10 10 10 10 

100 TE PaT ae 18) Tae pero MPs ALS) 79" 20 
10 =10 

the sum of the first and last or a= 55 + an 1,5 


9 


I22 ON THE RECTIFICATION AND 


the sum of the even erates o 


=7 +a e+e sta 3 io 3.459539 


The sum of all ie ber ee ordinates or € 


=< 10 
w+ iti at ~+= = 2. 72817445 


Therefore, by ‘the Bade 
(a + 4b + 2c)id=6,9315021, is the area required. 


PROPOSITION XVII. PROBLEM. 


To find the value of x by equi-distant ordinates to the revoloidat 
curve. 


If A = the area of a quadrant, 4_8 
AIR of the revoloidal eurve, anda | |S 
= the minor semi-axis, Al = the ra- 
dius of a circle inscribed in the curve, 
and 1 = the semi-major axis, IR ; 
and if @ = an ordinate EN to the 
major axis, equi-distant from I to the 
vertex R, then will the area of the 
curve, considering it as a parabola, 


be = a+4 bx453 and if A equal the area of the revoloidal 


uadrant, the semi-transverse = sal =< x = Les 1} 
4 Fain 1(q nae) — a+4ab (1) 
Let | a=1,and b= /1=70712, &c. 


and Alawiiltas foe 1 
P31 GB7D969 oe 
ELT ry ae ae 


this is true to two places of figures only, but the third should 
be 7 instead of 6. 

Let now two other ordinates, CL, GP, be taken in addition to 
this, equidistant therefrom, and from the extremities of the axis, 


3An t 
and we have by Prop. XVI, Cor. 2, rey Pa a i 


_ Let the two ordinates b be the sines of 22° 30’ = .382683, 
and 67° 30’ = .923880, and c = sine of 45° = .707107, and 
we have a= 1, 4a = 5.226252, 2c = 1.41424,A =r? = 1. 


u LN ne Taga Fa 1.5705848, é&c. = — 
ene 7 4b4- 20) 71640466 NP Toa 


which is true to four places, viz: 1,570, but the fifth should be 
7 instead of 5. 


hence we have 


QUADRATURE OF THE CIRCLE. 123 


Let, now, seven ordinates be taken between the two extremes, 
and the distance of the ordinates will be reduced . 

Thus the ordinates will be the sines of 11° 15’, 22° 30’, 33° 
45, 45°, 56° 15’, 67° 30’, 78° 45’, 90° = .195090. .382683. 
555570, .707107, .831470, .823880, .980785, 1. 


sin. 11° 15! = .195090 

__ jain, Of 0). 33° 45! = .555570 

G7 sin. GO° eatenh } ty 56°15’ = .831470 

78° 45’ = .980785 

2.562915 

sin. 22° 30’ = .382683 i 
C. = 45° 0’ = .707107 
67° 30’ = .923880 
20.13670 
Hence, a=1, 4)=10.251660, 2c=4027340, and n=8. 
3n T 
~ ° = ee) STP ete L = 
Therefore, in the formula FRC On eth nee have } « 


1.5707833, which is true to five places, but the sixth should be 
9 instead of 8. 
By comparing each of the results obtained above with the 
true numbers, we shall have the ratio of its approximation. 
Thus, the difference between the first result and the true 


number is, .0035724 
that of the second, .0002115 
that of the third .0000180. 


Hence, it will be seen that the result approximates in nearly 

a geometrical progression, to the true value of 7 as we in- 
crease the number of ordinates, or as the distance between 
the ordinates is decreased. We pay, therefore, determine the 
number of ordinates that must bé taken, in order to give an 
accurate result to any number Of decimal places ; for it will 
be perceived that the ratio of the above variations are nearly 
16 to 1. Hence, we may sfely infer, that it will approximate 
at the rate of three decimal figures in every two subdivisions. 
Instead of computing the value of } « from the ordinates drawn 
in the whole quadraxt, we may take any small arc of the 
quadrant, and having found its quadrature let it be called A’; 
and if we proceed’as before, by drawing one ordinate equi- 
distant from its extremes, we shall have, according to the for- 
1 / 

mula, tee = x’ = the assumed arc, which, multiplied by the 
number of times this arc is contained in the quadrant, will 


Se) 


124 ON THE RECTIFICATION AND 


give the same result as though the ordinates are taken for the 


whole quadrant. 


1A | 
Let the formula row; = «' be applied to a segment of the 


revoloidal surface whose arc is 30°, and whose greater ordi- 
nate is the radius. 
Here, A’=sin. 30° X r=}; hence, A’=14A, 
a=r + sin. 60° = 1.866025 
= sin. 75° = .965926 
6A! 3A ‘ " 
| BEES Ae OTTy =o siguagabampatiess ass PEE 


radius 1; hence the arc of 90°, or £7 = 1.57076010; 
which is true to five places of figures. 
Let there be two other ordinates drawn across the segment, 


3nA! 


——$—$___— = / ’ eee 18 Neieenae 
then by the formula, q44h42¢ 77? Since = gr+3, we shall 
9nA! 
have Arne ae Ir, and r= 4 


sin. 90° — .100000 = 1,866025 


fits 67° 30’ — .923880 
= ) sin. 82° 30’ — .991445 ( = 1.915825 


c== sin. 75° = .965926 


Hence, a+4b+2c = 11.459157 
and, 9nA’ = 18, 
9nA! 
AMD lak 1.57079616. 


which is true to seven places, which is as far as the sines are 
calculated, on which the value is predicated. 
Let the first and second results be compared, and we shall 
have the difference of the frst result and the true value, 
1.57079632 
—1.57076010 | = .00003622 
that of the second, 
1.5'7079632 
—1.57079616 | = -00090016 
Divide the first error by the last, aad we have the quotient 
= 226 = the ratio of the approximation, or the proportional 
accuracy of determination, by varying tne number of the or- 
dinates. It will be perceived, therefore, that this portion of 
the revoloidal curve approximates much faster than the whole 
quadrant, and is, therefore, more nearly simila: to a parabola 
than the whole curve ; it may hence be inferred, that if any 
small segment of the revoloidal surface is taken adjacent to 


QUADRATURE OF THE CIRCLE. 125 


the conjugate diameter, such segment will be very nearly a 
portion of a parabola, and that, by so considering it, the value 
of the length of the arc may be determined with any required 
degree of accuracy. 

For, in taking the whole quadrant, we found the ratio of 
convergency, by doubling the number of the ordinates, to be 1 
to 16; and in the segment embracing the arc of 30°, adjacent to 
the conjugate diameter, we find the ratio of convergency to be 
1 to 226; and if an arc is taken still smaller, the ratio of con- 
vergency will become proportionally greater. 

Let any segment of the revoloidal surface be taken, and if 
the value of its arc, or the value of +, be computed by any 
number of ordinates, and if the number of ordinates is then 
increased so that the common distance ‘is reduced one-half, 
and the value of 7 is again estimated by the increased number 
of ordinates, and if the variations of the two results from the 
true value be compared with each other, they will show the 
ratio of convergency of the process for that arc by increasing 
the number of its ordinates, or the rate of approximation by 
any specific increase of the ordinates for such arc or segment. 

Hence, we may at all times determine the value of 7 to any 
required degree of exactness ; for if we wish to determine its 
value to any given number of decimal places, we have only to 
assume some given arc and find its rate of convergency, then 
take such an arc as, according to this rate, will give the re- 
quired result. 

The arc of 90° gave the true result only to 2 places, that of 
30° to 5 places, with three ordinates; and we may expect a 
still greater ratio of convergence for a smaller arc; let us 
take an arc of 10°; we may, according to this ratio, only have 
the value to 8 places, and by proceeding to decrease the arc, 
we should, by taking 1°, have the value to 16 places; but 
since the curve approaches more and more to a similarity with 
that of a parabola as we approach the vertices of the conju- 
gate axis, the ratio of convergency increases also rapidly as 
we approach that point, or as the arc assumed is decreased ; 
so that, by taking an arc of one minute of a degree, the accu- 
racy of determination would extend to many places of deci- 


mals; andif the are should be reduced still further, to seconds | 


and fractions of a second, the result would come out true to 
several hundred decimal places; all of which is manifest by 
pursuing the investigation. 

Let a distance be taken on the axis equal the arc of 2 mi- 
nutes of a degree from the conjugate diameter. Then having 
the sine 89° 58’ = .9999998308 

that of 89° 59 = .9999999577 


= 


126 ON THE RECTIFICATION AND 


the cosine of 2 is .00058177637, which, multiplied by radius, is 
the value of A’ 
Wye have, in the formulae 
e BYE, in the formula 775 


6A’ = .00349065837 
a+ 4b = 5.9999996616 


= 0005817764172 = «’ = the arc of 20° 2’, which 


A’ 
a+4b 
multiplied by 2700, the number of such arcs contained in a 
quadrant, we have the value of 1 * = 1.57079632644, which 
is true to 10 places, or as far as the sines on which its value 
is predicated. 

For more extended investigations on this subject, see notes. 


and 


PROPOSITION XVIII. THEOREM. 


If a circle be described, and from its centre a line equal to one- 
fourth of the circumference be drawn perpendicular to a ra- 
dius, the triangle formed by connecting the extremity of this 
line with the extremity of the radius, will be equal to the quad- 
rant of the circle. 


Let ABE be a semicircle, described on the 
radius AC, from the centre C draw the line 
CD equal to one-fourth of the circumference 
perpendicular to the radius AC; join DA, 
then will the triangle ACD be equal to the 
quadrant AEC. 

For, according to Prop. XV, Cor. 1, B. 
V. El. Geom. the area of a sector of a cir- 8 C A 
cle is equal to half the product of the are of the sector mullti- 
plied by the radius. Now, the quadrant AEC is a sector of 
the circle, and the triangle ACD is equal to half the product 
of the arc AB, or the line CD multiplied by the radius AC; 
hence the triangle ACD is equal to the quadrant AEC. 


Cor. Hence we may infer that AF’, the segment of a circle 
cut off by the line DA, is equal to the portion of the triangle 
DEF cut off by the arc EF; for the triangle ADC is equal to 
the quadrant AEC, and if the line AD cuts off a segment AF 
from the quadrant, then it necessarily includes an equal space 
DEF within the triangle and without the quadrant ; otherwise 
the triangle ADC, could not be equal to the quadrant AEC. 


QUADRATURE OF THE CIRCLE. 127 


PROPOSITION XIX. THEQGREM. 


If a circle be described, and | from the centre two radial lines be 
drawn perpendicular to each other, equal w length te the are of the 
quadrant ef the circle; and if the extremities of these radial lines 
be connected with anether line as hypothenuse, forming with those 
dines a triangle, and if this third line is bisected by enother radial 
line from the centre, bisecting also the arc of the quadrant ; and if 
a line be drawn from the point of the hypothenuse cut by the last 
mentioned radical line, to the extremity of the radius, or the point 
where one of the _firstmentioned radial lines cut the circle ; then will 
the triangle formed by these three lastmentioned lines, be equal io 
the sector included between the two radial lines forming sides of this 
triangle ; and the lastmentioned line will cut off a segment from the 
‘sector without the triangle, equal to the portion included in the trian- 
gle without the sector, and a perpendicular let fall from the point of 
bisection of the hypothenuse on the radius, is equal in length to the 
arc of the circle included between the bisecting line and radius. 


From the centre C of the 
circle AB draw the radial 
lines CD and CG, each equal 
in length to the arc of the 
quadrant AE, forming a 
right angle at C, draw DG; 
draw also CH bisecting DG 
in H; and draw HA; then 
will the triangle ACH be 
equal to the sector ACT, and 
the segment Ab cut off from 
the sector by the line HA, 
will be equal to the portion 
HTd included in the triangle 
ACH, but without the sector ACT; from the point H of the 
intersection of the line CH with DG, draw HI perpendicular 
to AC, and the line HI will be equal to the arc AT included 
between the radial lines AC and CH. 

Draw the line AD forming with the radius CA, and the ra- 
dial line CD the triangle ACD ; and (Prop. XVUL) the triangle 
thus formed is equal to the quadrant AEC; draw also HI per- 
pendicular to CD; and because HL is parallel to AC, the two 
sides DG and DC will be cut proportionally by the line HL, 
(Prop. XIV, B. IV, El. Geom.) ; so that if the line DG is bi- 
sected in the centre at H, the line CD is also biseeted in the 
centre at L, so that LC or HI=!CD. 

Now, the two triangles ADC and AHC, having a common 
base, viz., AC, are as their altitudes, (Prop. VIII, Cor. B. IV 


128 ON THE RECTIFICATION AND 


iil. Geom.) ; but the altitude of the triangle AHC is HI, equal 
to LC, equal to half the altitude CD of the triangle ACD ; 
therefore, the triangle ACH is equal to half the triangle ACD; 
but the triangle ACD is equal to the quadrant ACE; therefore, 
the triangle ACH is equal to half that quadrant, or is equal to 
the sector ACT, which is the first branch of the proposition. 

Now, because the triangle ACH is equal the sector ACT, the 
segment Ab cut off from the sector by the line AH, is equal to 
the portion HT included in the triangle and without the sec- 
tor, (Prop. XVIII. Cor.) which is the second branch of the 
proposition. 

And because the line CD is equal to the are AE of the quad- 
rant, and because the arc AT is equal to half the are AX, it is 
also equal to half the line CD=CL=HI, which is the last 
branch of the proposition. 


Cor. If BA be extended to a, so that Ca shall be equal CH, 
and if a line Habe drawn and bisected by Ce, and a line be 
drawn from e to A, forming the triangle ACe, this triangle so 
formed will be equa] to the sector of the circle intercepted by 
the lines CA and Ce; and the line Ae will cut off a segment of 
the circle without the triangle equa! to the space included in’ 
the triangle without the sector, and a perpendicular ei let fall 
from e on the radius AC, is equal to the are intercepted by the 
lines CA and Ce. And the same may be inferred from any 
further divisions or subdivisions of the circle. 


PROPOSITION XX. THEOREM. 


With the radius CA let there be described a circle AEBF, and 
from the centre C draw the line CD perpendicular to 
the diameter AB, or radius CA, and equal in length to the 
arc of the quadrant AX, let the line CD be divided into any 
number of equal parts, as 1, 2, 3,4, &c., and let the arc of 
the quadrant be divided in like manner into a similar number 
of equal parts 1, 2,3, 4, &c. Krom the divisions on the line 
CD draw the lines la, 2b, 3c, &c., parallel to the radius CA ; 
and through the divisions 1, 2, 3, &c., on the are draw the 
radial lines Ca, Cb, Ce, Gc. Then, if from the points of in- 
tersection a, b, c, &c., of the radial lines, with their respec- 
tive parallel lines according to their respective numbers, lines 
be drawn as fA, to the extremity of the radius CA, then will 
this line, together with CA and Cf form atriangle which is 
equal to the sector CA6, included within the radial line Cfand 
the radius CA. 


QUADRATURE OF THE CIRCLE. 129 


And the segment Am cut off from the sector by the line fA is 
equal to the portion {m6 included in the triangle without the 
sector; and if lines be drawn from the several points of in- 
tersection of the parallel and radial lines perpendicular to the 
radius as {K, the lines so drawn will be respectively equal to 
the ares intercepted by the radial lines from the extremities 
of which they are drawn, and their radius CA. 


For each of the radial » 
lines Ca, Cb, &c., cut the 
arc AE in the same ratio 
that the corresponding pa- 
rallel lines la 2b, &c., cut ar 
the perpendicular CD ; thus ww} ° 
the radial line Cg, passing 2% 
through the point or divi- lf] 
sion 7 on the arc, cuts off _«/,7/ 
one division from the arc, 
and intercepts with the ra- 
dius.CA all the rest, and the 
corresponding line 7g pa- 
rallel to the radius CA cuts 
off D7 on the perpendicu- 
lar CD, so that if the whole 
line CD is equal to the are ¥ 
AE, the portion C7 of the line CD = Hgis equal to the are 
A7 of the circle. And the radial line C6f cuts the arc AE in 
the same ratio as the line {6 parallel to CA, cuts the line CD, 
viz., the radial line C6f cuts the circle through the division 
marked 6, and the line f6 parallel to CA, cuts the line CD in 
the division marked 6, so that if the whole line CD equal the 
arc AE, then will the portion C6 of that line= fIK be equal to 
the portion A6 of the are. Now, the area of every sector of 
the arc AC6, is equal to the arc of the sector multiplied by half 
the radius CA ; but the triangle AfC is equal to the line fK 
_ multiplied by half the base or radius CA ; therefore, ‘the 
triangle AfC is equal to the sector AC6, and hence the seg- 
ment Am, cut off from the sector by the line Af, is equal to the 
portion fm6 included within the triangle, but without the circle, 
and, as has been shown, fK=C6 is equalto the arc A6. And 
as the same holds true in each of the points of intersection a, 
b, c, d, &c., it follows that the result corroborates the affirma- 
tion expressed in the proposition. 


Cor. If a curve line be drawn through the several points a,b, 
c,d, &c., anda triangle be formed by, two lines from any point 
in the curve drawn to the two extremities of the radius, the 


130 ON THE RECTIFICATION AND 


triangle so formed, will be equal to the sector included between 
the radius and the other line terminating in C, and the same 
relation of areas and lines will exist with regard to the trian- 
gles and lines drawn from any point of the curve, as though 
they were drawn through the points a, b, c, d, &. 


Scholium 1. The curve BD may be described about the 
quadrant BE in a similar manner; and since from any point 
in this curve, if a triangle is constructed on the radius as a 
base, this triangle is equal to the sector of the circle included 
between the radius and one of its sides, the curve may be 
called the curve of the circle’s quadrature. This curve varies 
from the revoloidal curve, inasmuch as the revoloidal curve is 
formed by drawing a line through the points of intersection of 
a series of lines parallel to the radius drawn through their re- 
spective divisions on the perpendicular, with an equal series 
of lines perpendicular to the former, drawn through their re- 
spective divisions on the arc (see Prop. II.) 


Scholium 2. This curve is generated a‘ 
by the locus of the intersection Q. of the £ 
two right lines CF,HG; HG being made 
to pass uniformly along from A to C, 
being always perpendicular to AC, while «6 
CF revolves about the centre through 
the arc ED. 

Let the origin be at A. 

Let AH |= 2, HQ = y 

AC = DE = ,' angle ACQ =4, tS = 

sine 6 =s, tM = cosine 6= cc 

Then c:s::9/—a:y E m1 c 

Sr'-—S$x i 
Hence y = tah which is the equation to the curve. 


ot 


PROPOSITION XXI. THEOREM. 


Uf from the extremity of the radius of a circle, a chord be drawn 
cutting off any segment less than a semi-circle, and if from 
the centre of the circle a secant be drawn through the opposite 
extremity of the segment, and if the secant be produced so that 
a line drawn from iis extremity, perpendicular, to meet the 
diameter produced, shall be equal to the arc of the seg- 
ment, then the area of the segment will be equal to that 
of a triangle formed by the chord of the segment, and the 
part of this secant line without the circle, and aline joining 
the opposite extremities of this line with that of the chord. 


QUADRATURE OF THE CIRCLE. 131 


For it is evident from the converse of Prop. X VIII Cor., that 
if CE be so drawn that the perpendicular Ez shall be equal to 
the arc Aa, then the triangle CEA will be equivalent to the 
quadrant AaC ; take away the triangle ACa, then there will re- 
main the triangle Aak = the segment Afa. 

First, let Aa be a chord cutting off the 
segment Afa ; from the centre C through 
the extremity of the segment at a, drawa 
secant line Ca produced to E, so that the 
perpendicular Et on the diameter, shall 
be equal to the arc Afa cut off by such 
secant ; then will the area of the seg- 
ment Afa be equal to that of the trian- 
gle AaE formed by the chord Aa, with 
the part aE of the secant without the circle, and the line AE 
joining the opposite extremities. 

Secondly, let the chord Ab extend E 
into the second quadrant, cutting off 
the segment AD#, draw Cbd, and ex- 
tend it to,E, so that the perpendicu- 
Jar EF drawn from the point E to the 
diameter AB, produced, shall be 
equal to the arc ADod cut off by the 
chord Ad, or the secant CE, and the 
area of the segment ADb will be 
equal to that of the triangle ADE 
formed by the chord Ad, the part of 
the secant 0H, and the line AE join- 
ing their opposite extremities. 

For, in the triangle ACE, the area is equal to the base AC, 
the radius of the circle multiplied by half the altitude EF, or 
the arc ADb of the segment, but the sector ADOC is 
equal to the radius AC multiplied by half the are ADb ; hence 
the triangle ACK: = the sector ADOC, therefore, if we take the 
triangle ACd from each, we shall have the segment ADb = 
the triangle ADE. 


PROPOSITION XXII. THEOREM. 


If from the extremity of the radius of a circle, a chord be 
drawn cutting off a segment greater thana semi-circle, and 
if through the opposite extr emit y of the segment, a secant be 
drawn from the centre of the circle, and if the secant be 
produced, so that a line be drawn -from its extremity on the 
diameter, produced, if necessary, shall be equal to the arc of the 
segment, then the area of the segment will be equivalent to 
the triangle formed by the radius, secant line, and a line 
joining the opposite extremities of these lines, plus a tri- 


132 ON THE RECTIFICATION AND 


angle formed by the radius and the chord, with the hne- 
joining the opposite extremities of those lines. 


For, let the chord extend into 
the third quadrant, cutting off the 
segment ADBdA, greater than a 
semi circle, draw Cd, and produce 
it to E, so that the perpendicu- 
Jar EF, on the diameter AB pro- 
duced, shall be equal to the arc 
ADBd of the segment, and join 
AK, and the area of the triangle 
ACE, plus ACd will be equal to 
the segment ADBdA. 

For the area of the triangle 
ACE is=the base AC, or radius, 
multiplied by half the perpendicu- 
Jar FA, which is equal to the arc 
ADBd, by hypothesis ; but the 
area of the sector ADBdCA is 
equal to the radius CA multiplied 
by half the arc ADBd, hence the q 
triangle ACE = the sector ADBdCA. — Add to both the tri- 
angle ACd, and we have the segment ADBCA = the triangle 
ACE + the triangle ACd. 

Secondly, let the chord 
Ae extend into the fourth 
quadrant. cutting off the seg- 
ment ADBFE. Through the 
point « draw the line CE 
produced, so that the line EF 
perpendicular to BA produc- 
ed, shall be equal to the arc 
ADBFe. Then the segment 
ADBFe will be equal to the 
trian. ACE + the trian. AeC. 

For the area of the trian- 
gle ACE is equal to the base 
AC, or radius multiplied by 
half the perpendicular FE, 
which is, by hypothesis,* 
equal the arc ADBF ; but the. 
area of the sector CADBFeC 
is equal to the radius AC 
multiplied by half the are |// 
ADBFe ; hence the triangle 
ACE is equal the sector © 


E 


A 


QUADRATURE OF THE CIRCLE. 133 


CADBFeC. Add to each the triangle ACe, and we have the 
segment ADBFe = the triangle ACE + the triangle ACe. 


Scholium. It may be observed, that, as the termination of the 
segment approaches the point B, or as the segment becomes 
equal to the semi-circle, its equivalent triangle becomes infi- 
nitely extended in the line Ali, and at the same time the sine 
becomes infinitely small, and while it passes the point B, the 
sine is equal to 0, and AE is infinite. The same may be said 
as it approaches the point A on the fourth quadrant, and be- 
comes equal to a complete circle. 


PROPOSITION XXIII. THEOREM. 


The area of a segment of a circle is equal to half the product 
of the difference between the arc of the segment and its sine 
multiplied by the radius. 


First, let Afa (see first diagram to Prop. X XI) be a segment 
of a circle cut off by the radial line Ca, and produce it to E, 
so that the perpendicular Ez on the radius will be equal to the 
arc of the segment ; join EA and (Prop. X XI.) the area of the 
triangle AEKa will be equal to the segment Afa ; from the point 
a draw the perpendicular as, which is the sine of the arc Aa 
of the segment, and the area of the triangle AKa will be equal 
to (Et—as) !AC (Prop. XX XI, B.1V, Hl. Geom.) Hence, the 
segment being equal to the triangle, A Ea is equal to half the pro- 


duct of the difference of the arc, and its sine multiplied by the 
radius. 


Secondly, let the chord Ab (see second diagram to Prop. X XI) 
extend into the second quadrant, draw CE, making EF = the 
arc ADb; draw AH, and the triangle AEKb will be equal to the 
segment ADd; draw bs, the sine of the arc ADb of the seg- 
ment ; then will the area of their triangle AKd be equal to 
' (EF -- bs) AC; (Proposition XXXI, Book IV, EHlements 
Geom.) ; hence the segment, being equal to the triangle, is 
equal to half the difference of the arc and its sine multiplied by 
the radius. 

Thirdly, let the chord Ad (see first diagram to last prop.) 
extend into the third quadrant, cutting off the segment ADBd, 
greater than a semi-circle. 

Draw Cd and extend it to KE, making EF equal to the arc 
ADBd ; join IA, and the area of the segment ADBd will be 
equal to the triangle ACK+ACd, (Prop. XXII) ; draw ds per- 
pendicular to the radius CB, and this line will be the sine of the 
arc ADBd ; and since it is a sine of an arc greater than a 


semi-circle, its value is to be considered negative, by Trigono- 
metry. 


ov 


134 ON THE RECTIFICATION AND 


From EF substract — dS, and we have EF + dS, which. 
if we multiply by AC, will give the area of the triangle AEC 
+ACd, which (Prop. XII.) is also equal to the segment ADBd. 
Hence the area of the segment ADBd is equal to the differ- 
ence of the arc, and its sine multiplied by the radius. 

Fourthly, let the chord Ae (see 2d fig. last prop.) extend into 
the fourth quadrant, cutting off the segment ADBFe; draw EC 
so that EF shall be equal tothearc ADBF¢e of the segment joinEA, 
and the area of the segment ADBFe will be equal to that of 
the triangle ACE + the triangle ACe (Prop. XII); draw es, 
the sine of the arc ADBFe, which, by Trigonometry, is nega- 
tive, being below the diameter, and in the fourth quadrant, 
which sine substract from the line EF, and we have EF + es, 
which, multiplied by half AC, gives the value of the segment 
ADBFe. Hence we have, as before, the segment equal to half 
the product of the difference of the are of the segment, and 
its sine multiplied by the radius. 


Hence we may infer, that a circular zone or a portion of 
the circle included between two segments is equal to half the 
product of the difference of the arc of the zone, and one of 
the segments included, and the sine of such arc, — the dif- 
ference between the arc of the included segment, and its 
sine multiplied by the radius. 

For if ABDE bea zone, and ABL L 
be a segment of the circle C, the seg- 
ment EALBD equal to the zone ABDE, 


and segment ABL; but these segments is 
are respectively equal to the excess of rs 

their several arcs above the sines mul- . 
tiplied by half the radius. Therefore, E D 


the zone ABDFE, is equal to the differ- 
ence of the excess of the arcs EALBD, 
and ALB, above their respective sines, multiplied by half 
the radius. The same may. be shown in relation to the 
area ABDFE in reference to its external segment EDF, or 
AEFDB, regard being had to the positive and negative va- 
lue of the sines. : 


r 


Scholium. Let A represent the area of the segment of a cir- 
cle ; let « = the arc of the segment, and s = sine of that arc. 
and let 7 = radius, and the segment may be expressed as in 
the following formula 


=1(re— rs) - - “ (1) 
Ifr = 1, then the expression becomes 
A = 1 —s) it ms ‘ (2) 


whence the segment is expressed in terms of the arc and sine 


QUADRATURE OF THE CIRCLE. 135 


Let A’ = the area ofacircular zone ABDE 
and x’ = the are EKALBD 
s' = the sine of that arc. 
Then will the area of the zone be expressed by ' 
A'=1(re'—r1s')—7(727—1s)_- - : - (3) 
If r=1 A’=}(a'---s')—--}(a---s) - 4 - : . (4) 

The arc may be expressed in terms of the numbers represent- 
ing the segment, and the sine of the arc, by the formula, « = 
pd ag tt : - - - - - : > (5) 

And the sine, in terms of the arc and segment by 

BEV s Se) DIA: - - - (6 

Sch. That the area of the segment of 5) 

a circle is equal to half the difference of 

the arc of the segment, and its sine X by 

the radius may also beshown as follows. 4 B 
Let AE be the segment of a circle, ats. 

whose radius is AC, draw ES perpendi- 

cular to the radius, and ES will repre- 

sent the sine of the arc AE; let AC be 

represented by r, and ES by s, and the arc AE by 2, then will 
3 re = the sector ACE, and irs = the triangle AEC, and 
+ rx —trs = the difference of the sector and triangle = the 
segment AK, viz., the seg. is = half the difference of the arc, 
and its sine multiplied by the radius. 

Also, in the segment EFB let CB, be represented by r, and 
the arc EFB by 2, and ES will be the sine of the arc EFB, 
1 ra = the sector ECBF, and irs = the triangle ECB. irz 
—zrs = their difference = the segment EBF. 


PROPOSITION XXIV. THEOREM. 


The area of the space intercepted by a tangent and secant with- 
out a circle, is equal to half the product of the difference of 
the tangent and arc, intercepted by the secant, multiplied by 
the radius. 


Let ATF be the space intercepted by 
the tangent AT, and the secant CT, with- 
out the circle; and the area of that space 
will be equal to half the product of the 
difference of AT and the arc AF, multi- 
plied by the radius AC. ? 

For the arc of the triangle ATC is equa 
to ;AT x AC, and the area of the sector 
ACF is equal to half the arc AFXx AC. 


Let t=the tangent AT, and «=the arc AF, r=AC. 


# 
r 
136 ON THE RECTIFICATION AND 


Then we have for the triangle ACT, rt, and for the sec 
tor AFC, 1rz; if from the triangle we take the sector, we 
have Lrt—1rz=the space AFT, hence as in proposition. 


Scholium. Draw Fs, FA, and Fv; then Fs will represent the 
sine of the angle ACF’, and sA or Tv, the versed sine; hence the 
traingle ATF equals one-half the product of the differenee of the 
tangent and sine multiplied by radius equals one-half the pro- 
duct of the tangent and versed sine ;-draw As’ perpendicular 
to CF, and this will also represent the sine of the angle ACF 
on the radius CF; hence, also, the triangle ATF is equal to 
one-half the difference of the secant CT and radius multiplied 
by the sine. 


PROPOSITION XXV. THEOREM. 


If about a plane revoloidal surface from a quadrangular revo- 
loid, a square be described, the space enclosed by the square, 
but without the revoloidal surface, will be to that contained by 
the square, as the sum of an infinite series of segments of the 
quadrant of the circle, whose arcs are in arithmetical pro- 
gression, to the sum of a similar series of sectors subtended 
by the same arcs. ae 


Let ACD be a quadrant of a plane 
revoloidal surface, and let HCD be 
a triangle forming one-fourth of a 
square circumscribing the whole re- 
voloidal plane surface. Divide the 
axis CD into any number of equal 
parts as 1, 2, 3, 4, &c., and through 
the points of division draw ordinates 72 
la, 2), 3c, &c., parallel to HC, and g ZAM ee 
these ordinates will cut the surface YZ fi 
of the quadrant of the revoloid and H A Cc. 
its circumscribing triangle, in the relations of their magnitudes, 
through the portions where such ordinates pass; and if the 
number of those ordinates be indefinitely increased, the sum 
of those drawn across the triangle HCD, will be to the sum 
of those drawn through the revoloidal surface ADC, as the 
area of the triangle to the area of the revoloidal surface. Now, 
the ordinates al, 62, c3, &c., are equal to the arcs represent- 
ed by numbers on the arc AF of the quadrant, corresponding 
to those on the axis, (Prop. II, and Cors.) ; and the portions 
17, 16, m5, &c., of those ordinates, are severally equal to the 
sines of those arcs, (Prop. XIII, Cor. 1,); hence the portions of 
those ordinates, intercepted by the curve AD and line DH. are 


5 


QUADRATURE OF THE CIRCLE. 137 


severally equal to the difference of the arcs and sines repre- 
sented by these ordinates. And since we have shown (Prop. 
XXIII.) that the area of the segment of a circle is equal to half 
the difference of the arc of the segment and its sine multiplied 
by radius, and since the area of the sector of a circle is equal 
to half the arc of the sector multiplied by radius, it follows 
that the sector and segment containing the same arc, are to 
each other, as the arc is to the difference of the arc and sine. 
Let a equal the arc, s equal the sine, r equal the radius. 

Then  tra=the sector, 
and tra— $rs=the segment containing the same arc, which 
are to each other in the ratio of , 

tra: 4ra— Grs, or of a: a—s. 

And since this is true for a segment and sector, contained 
by any arc in that quadrant, that is, the sector of a circle is to 
the segment containing the same arc as the arc is to the arc mi- 
nus the sine, or as the ordinates g7 to gi, or as f6 : f 1, &c.; and 
as this relation evidently exists in reference to each of the par- 
allel ordinates, and as the ordinates represent arcs of the cir- 
cle in arithmetical progression, it follows that those ordinates 
drawn across the triangle, may also represent a series of sec- 
tors of a circle, while the portion of those ordinates intercept- 
ed by the curve AD and line HD, may represent a similar se- 
ries of segments of the circle. And as we have shown above 
that the surfaces HCD and HAD are to each other in the re- 
lation of the ordinates passing through each, it follows that 
the trilinear space ADH will be to the triangle HCD as the 
sum of an infinite series of segments of the circle whose arcs 
are in arithmetical progression, whose first term is equal to 
the common difference, and whose last term is the quadrant of 
the circumference to the sum of a similar series of sectors 
with the same arcs. And since the whole plane revoloidal 
surface consists of four quadrants, ADC, and its circumscrib- 
ing square consists of four triangles, HCD, it follows ‘that the 
whole revoloidal surface will be to the whole circumscribing 
square in the same ratio. 


Cor. The revoloidal quadrant ACD, may be represented by 
a similar series of isosceles triangles formed by the chords of 
the series of segments with two radii drawn from the extremi- 
ties of those chords. For we have shown that the series of 
sectors may be represented by the area HCD, while the series 
of segments are represented by HAD; if we take the seg- 
ments from the sectors, we shall have the triangles, evidently 
equal HCD — HAD=ACD. 


10 


", ¢ 


138 ON THE RECTIFICATION AND 
PROPOSITION XXVI. THEOREM. 


The area of a quadrangular revoloidal surface, from a right 
quadrangular revoloid, is to that of its circumsribing square, 
as the sum of an infinite series of sines of the quadrant whose 
arcs are taken in arithmetical progression, and whose com- 
mon difference is equal to.the first term, to a similar series of 
ares of such sines. 


v 


Let DHEI be a square circumscribing the plane revoloidal 
surface DAEB, and let ADC be a quadrant of that surface, the 
triangle HDC being its corresponding portion of the square ; 
divide the semi-axis CD into any number of equal parts, as 1, 
2, 3, 4, &c., and through the points of division draw the ordi- 
nates la, 2b, 3c, &c., parallel to CH, those lines will each be 
equal to such arc of the quadrant AF as corresponds to the 


E 

point of division on the line CD, in reference to similar divi- 
sions of the quadrant AF’; thus the line 1a equal the arc F1 on 
the quadrant, since it is evidently equal to 1D on the axis CD; 
and (Prop. II, Cor.) 1D=1F; also, if we take the line 2b=2D, 
this is also, for similar reasons, equal to the arc 2F; and hence 
each line parallel to CH, terminated by the lines CD, HD, are 
equal to the arc, corresponding to the divisions on the line CD, 
through which it passes. 

Again, the portion of these lines or ordinates intercepted by 
the axis CD, and the curve AD, (Prop. X, Cor.) are severally 
equal to the sines of the arcs, corresponding to the divisions 
from which they are drawn on the axis. And since, by hypo- 
thesis, the arcs are taken in arithmetical progression, they 


QUADRATURE OF THE CIRCLE. 139 


must be equidistant; hence, if the number of such ordinates 
are indefinitely increased, the sum of the portions of them in- 
tercepted by the axis CD, will represent the area or the sur- 
face ADC, as the whole ordinates drawn across the triangle 
HDC represent the area HDC. Therefore the surface ADC is 
to the surface HDC as the sum of the series of sines, to the 
sum of a similar series of arcs taken as above. Now, since 
this is the case with one quadrant, it must also be true in rela- 
tion to the whole revoloidal surface and its circumscribing 
square. 

Cor. 1. Hence, if a square KLMN be described on the di- 
ameter of the inscribed circle, the square so described will be 
to the square DHEI as the sum of the series of sines of the 
quadrant, to the sum of a similar series of arcs of the sines ; 
since the square KLMN (Prop. III, B. II,) is equal to the 
area of the revoloidal surface DAEB. 


Cor. 2. Each of the ordinates al, 62, c3, &c., drawn across 
the triangle HCD parallel to HC, is equal to the are of the cir 
cle represented by the corresponding numbers in the divisions 
of the quadrant AF. Thus, al equal the are 1F, 62 equal the 
arc 2F, and g7 equa) the arc 7F or the quadrant. 


ON SPIRALS. ; 


There is one class of curves which are called spirals, from 
their peculiar twisting form. They were inven by the an- 
cient geometricians, and were much used in architectural or- 
naments. Of these curves, the most important as well as the 
most simple, is the spiral invented by the celebrated Archi- 
medes. 

This spiral is thus generat- 
ed: Let a straight line SP of 
an indefinite length move uni- 
formly round a fixed point 8, 
and from a fixed line SX, and 
Jet a point P move uniformly 
also along the line SP, start- 
ing from 8, at the same time 
that the line SP commences 
its motion from SX, then the 
point will evidently trace out 
a curve line SPQRA, com- 
mencing at 8, and gradually 
extending further from S. 


146 ON THE QUADRATURE OF CURVES. 


When the line SP has made one revolution, P will have got to 
a certain point A, and SP still continuing to turn as before, we 
shall have the curve proceeding on regularly through a series 
of turnings, and extending further from S. 

To examine the form and properties of this curve, we must 
express this method of generation by means of an equation 
between polar ordinates. 

Let SP=r7r,SA= 6, ASP = @; 
then since the increase of r and éis uniform, we have 
SP:SA:: angle ASP: four right angles: : 6 : 20 


bd b 


ate — 6 j — 
Bic a, if a “e 

From this equation it appears that when SP has made two 
revolutions or 6== 47, we have r = 2b, or the curve cuts 


the axis SX again at a distance, 2SA ; and similarly after 
3, 4,7 revolutions it meets the axis SX at distances 3 ; 4,7 
times SA. 

Let any number of concentric circles be described, whose 
radii IA, IC are in arithmetical progression, and if the cir- 
cumference of the outer circle is divided into any number of 
equal parts, and radii are drawn from each of the points of 
division in the circumference of the circle to the centre I, 
lines drawn from the centre in such manner as to pass through 
the points of intersections, of the several radii with the curves 
in consecutive order, will be spirals similar to those of 
Archimedes. 


If the curve line pass from ete 
I, through rpB, or through A EST 
rgA, the curves IrgA, IrpB KE so 4S 
will be spirals; so also the =f JSC Aga TAS, 
lines IsdmD, IsdnD, and IwhfC A Z PE TeX ce 
and spirals; all generated by / / (FLEES ‘ 
the same laws, but with differ- AGA ales Nal LN B 
ent ratios of their angular, { \ \\' xt LSPA Phi 
compared withtheir rectilenear (CY) LHe 
motion ; or their circular, with \, Vat Se) “anh ff 
their radial motion. The two Gaps pee SPY 
spirals IsdmD, IsdnD, com- ne ee a ie 
mencing on the line IC, and De 


terminating at D, form the heart-like figure IsdmDndsl. 


These spirals, and especially the one with a heart-like form, 
are extensively used in mechanical operations ; to communi- 
cate a uniform rectilenear, reciprocating, from a rotary motion; 
it is therefore important that they receive some consideration. 


- 


ON THE QUADRATURE OF CURVES. 141 


The area of any portion included between the spiral and 
its circumscribing arc of the circumference terminated by the 
radius, its origin, is equal to two-thirds the sector, having the 
same arc of the circumference as its base. 

For let fa, Ib, Ic, &c., be the radii of circles in arithmetical 
progression, then will the arcs of these circles intercepted by 
two radii, be also in arithmetical progression, and since the 
value of 4 also increases uniformly in arithmetical progression 
along with r or , hence the value of the intercepted parts of 
the several arcs will be a series of arithmeticals multiplied 
into another corresponding series of arithmeticals, therefore 
their products will be a series, of the squares or a series of 
numbers proportional to a series of squares of a series of arith- 
meticals. 

It has been shown (Prop. IV, Cor. 3, B. I) that the sum ofan 
infinite series of the square of a series of numbers in arith- 
metical progression increasing from 0, is equal to 1 of the last 
term multiplied by the number of terms. 

But the arc intercepted by the spiral with the radius, as its 
origin is‘ the last term, and the radius represents the num- 
ber of terms ; hence the area IrpBCI is equal to 1 of the pro- 
duct of the are BC x IC; and the area IsdmDBCI is = 4 
arc CBD xX IC; also the area IvhfCADBCI is =} circumfer- 
ence ACBD x IC. But the area of the whole sector IBC, 
ICBD, or ICBDAC in either case is equal to half the product 
of their respective arcs, multiplied by the radius; hence the 
space intercepted by the spiral in each case is } that of their 
respective sections. 


Cor. 1. Hence the space IrpBI, is =} the sector ICB; 
the area IsdmDI is = + the sector ICBD or } the semi-cir- 
cle ; and the area IvAfCI is = 4 of the whole circle, 

Also the heart IsdmDndsI = 1 of the whole circle. 


Scholium 1. The spiral of Archimedes is sometimes used 
for the volutes of the capitals of columns, and in that case the 
following description by points is useful. (See first diagram. 

Let a circle ABCD be described on the diameter CSA, an 
draw the diameter BD at right angles to AC ; divide the ra- 
dius SC into four equal parts, and inSB take SP=1SC, in SA 
take SQ='SC, and in SD take SR=2SC; then from the equa- 
tion to the curve these points belong to the spiral ; by subdi- 
viding the radius SC and the angles in each quadrant we 
may obtain other points as in the figure. In order to com- 
plete the raised part in the volute, another spiral commen- 
ces from SB. 


142 ON THE QUADRATURE OF CURVES. 


§ 

Scholium. These spirals are of the same kind, as those 
formed by winding achord around a conical spire, from the 
vertex to the base, in such manner, as to encircle the spire at 

qual distances ; the exact length of such curve is of difficult 
determination. 

. The same spiral would be represented by the convolutions 
of a conical screw ; also, by a screw represented on a disc. 
Ifthe origin of the spiral is C 
f any point M, not in the cen- 

re of the concentric circles, 
then the area AFCM123A 
between the spiral and the 
outer circumference is=1 of the , 
product of the are ACB through 
which the curve would have 
passed from the centre I, mul- 
tiplied by the radius — } of the 
arc LMxIM — } (are LM + 
arc BC) x MC. 


D 
If PQM be a triangle, whose base PQ = the semi-circum- 
ference ACB = the angular space passed through by the two 
spirals MA, MB, then either portion PCM, QCM of the tri- 
angle may be expressed by } PC or 1 ACXCM = ! arc CBx 
IC—; arc LMxIM—t(arc LM + are BC) x MC, from this 


P Cc Q 


- 
substract the expression for the area included within the spiral, 
and the arc AB, and we have? ACB x IC—_1LM x IM= 
the difference of the areas. 


ON THE CYCLOID. — 148 
THE CYCLOID. 


if a circle EPF be made to roll in a given plane upon a 
straight line BCD, the point in the circumference which was 
in contact with B at the commencement of the motion, will, in 
a revolution of the circle, describe a curve BPAD, which is 
called the cycloid. 

This is the curve which a nail in the rim of a carriage-wheel 
describes during the motion of the carriage on a level road. 
The curve derives its name from two Greek words signifying 
* circle formed.” 

The line BD which the circle passes over in one revolution 
is called the base of the cycloid ;if AQC be the position of the 
generating circle in the middle of its course, Ais called the 
vertex and AC the axis of the curve. The description of the 
curve shows that the line BD is equal to the circumference of 
the circle, and that BC is equal to half that circumference. 
Hence also if EPF be the position of the generating circle, 
and P the generating point, then every point in the circular 
arc PF, having coincided with BF, we have the line BF = the 


arc PF, and FC = thearc EP or CQ; 


hs QM parallel to the base BD. 
Let A be the origin of the rectangular axes, 
AC the axis of x, and O the centre of the circle AQC. 
Fer AMi== 2, AU= a, 
MP =y, angle AOQ = 4: 
ae by the similarity of the position of the two circles, we 
ave 
PN = QM, ond PQ = NM ; 
~. MP= PQ + QM = NM+ QM= FC + QM = arc CQ 
+ QM that is, y = ad + asin. 6 = a (6 + sin. 4) (1) 
x= a@—acos. 6 = avers. 4 - (2) 
The equation between y and zis found by eliminating 4 be- 
tween (1) and (2) | 


al : 


4 


144 ON THE CYCLOID. 


a—x /2 axr—x? 
cos. 6 = 2 Sine = 
a a 
andy = aé-+ asin, 4 
—liqe-- x 
= acos. ( z + V¥ ax—x? 


But we can obtain an equation between x and y from (1) 
alone ; that is from the equation, AP = are CQ + QM. 

For arc CQ = a circular arc whose radius is a and versed 
sine x | 


° z i 
= Wes 4 a circular arc whose radius is unity and vers. sin. tix 


—1! 
a 
—l1 


x mst rie ey 
. Y =a vers. ie; +/9 a aa? 


If the origin is at B, BR = a and RP = y, the equations are 
«= aé---a@ sin. 4 
= a--- a cos. 4, 

We shall not discuss these equations at length, as the me- 
chanical description of the curve sufficiently indicates its 
form. 

The cycloid, if not first imagined by Galileo, was first ex- 
amined by him ; and it is remarkable for having occupied the 
attention of the most eminent mathematicians of the seven- 
teenth century. 

Of the many properties of thiscurve the most curious are, that 
the whole area is three times that of the generating circle, that 
the arc CP is double of the chord of CQ, and that the tangent 
at P is parallel to the same chord. . Also that if the figure be 
inverted, a body will fall from any point P on the curve to the 
lowest point Cin the same time; and ifa body falls from one 
point to another point, not in the same vertical line, its path 
of quickest descent is not the straight line joining the two 
points, but the arc of a cycloid, the concavity or hollow side 
being placed upwards. 


BOOK. V. 


ON THE PRODUCTION AND RESOLUTION OF GEOMETRICAL 
MAGNITUDES, CONSIDERED AS LINES, SURFACES, AND 
SOLIDS, EXISTING IN THEIR SPECIFIC RELA- 
TIONS OF FORM AND PROPORTIONS. 


CHAPTER 1. 


DEFINITIONS AND PRINCIPLES. 


Art. 1. We have hitherto referred lines, surfaces and solids, 
in all their varieties of figures and species, to some specific 
quantities and relations which were cognizable in such mag- 
nitudes, and whose properties were rendered evident to our 
consideration. Magnitude we have compared with magni- 
tude; figure with figure; and we have thereby established 
their relations, under arbitrary considerations. 

We will now consider magnitudes in the relation of their 
organization, or in the relation of their laws of production ; 
and instead of referring magnitudes to specific magnitudes ar- 
bitrarily chosen, we will refer them to others, only in the rela- 
tion of their laws of generation. 

2. Since a point by definition is locality without extension, 
any number of associated points cannot possess magnitude, 
hence a magnitude is not a multiple of one, or any number of 
points. 

3. Neither can any number of lines, however associated, 
constitute a surface, since lines are supposed to possess no 
breadth or thickness, one of which is essential to a surface; 
for if one line does not possess breadth, neither can any num- 
ber of associated lines; and if a line be multiplied by any ab- 
stract number, since it is expressed only in relation to its length, 
it can only be multiplied or increased in that relation. 

4. So, also, if a surface be multiplied by any number, in 
itself considered, the product cannot be a solid; for since the 
surface possesses no thickness, it does not possess the charac- 
teristic of the solid, and hence any number of such surfaces, 
or multiple of such surfaces in themselves considered, cannot 
be a solid. 

5. The distance between any two points is a line. For a 
point being locality without extension, if there be two locali- 


146 PRODUCTION AND RESOLUTION OF 


ties, they must be seperate from each other, and their distance 
from each other is necessarily extension in space, which agrees 
with our definition of a line, viz., “extension in one dimension.” 

6. Space is a medium in which all positive objects, and all 
local relations exist ; its existence is only indicated by its uni- 
versal property of extension; it is infinitely divisible in each 
or all its three dimensions of extensions, and infinitely exten- 
sible. 

7. Any definite portion of space, or any extension in space, 
is magnitude. 

Magnitude may possess extension in one, two, or three di- 
mensions, but space can properly exist only in its three dimen- 
sions of extension ; if it can be divested of extension in one 
dimension, it can in another, and so on till its extension is ex- 
tinguished. Magnitude may be properly applied to extension 
in whatever degree it exists; but space cannot properly exist 
independent of its three dimensions, which are its essential 
properties. ! 

8. If there be two points A, B, the first point A, drawn 
through the distance AB, produces or, describes 
the line AB; that is, the distance from AtoB <A B 
in the portion of space passed through by the 
point A, or the locality occupied by the point A in its passage. 
is the line AB. | 

9. If a line be moved through any space in.a direction not 
agreeing with its length or extension, the locality passed 
through by the line is a surface. 

Thus, if a line AB, be moved from its position AB to CD, it 
will, by that means, generate the surface ABDC, for the line 
will have occupied every portion of the extension between the 
two lines AB and CD, which cannot be said of any limited 
number of lines placed in juxtaposition across the figure ABDC. 

If, in this motion, the line AB always mainiaiis its parallel 
position, and if any point A ©) Acommmee OAT | eC 
in the line, describes a right 4 i i | 
line AC, the surface will be a 
a rhomboid or parallelo- | 
gram; and if, in addition to 
this, the line AC, or the di- pe D 
rection of its motion is per- |B 
pendicular to AB, then will the rhomboid be a rectangle. 

But if the line AB in its motion should not preserve its par- 
allel position, or if the distance BD, passed through by the point 
B, is greater than AC, then the figure generated will depend on 
the nature of the lines which serve as its boundaries, but in 
general, in such case, one or both of the lines AC, BD will be 


a 


GEOMETRICAL MAGNITUDES. 147 


curves, the nature of which will depend on the specific varia- 
tion of the course and inclination of the generating line. 

If instead of one line AB being drawn through the whole 
distane AC, a series of parallel lines AB, ab, &c., are drawn 
through their respective distances from each other, the result 
will be similar; or the same surface will be described by the 
series drawn through their several small distances, as by the 
single line drawn through the greater distance. 

The geometrical operation of drawing a line through a given 
distance to produce a surface, is equivalent ‘to that of multi- 
plying a line by a line, the product of which we have shown 
in the elements of geometry to be a surface. 

The measure of the surface generated by the motion of a 
line, is the length of the line multiplied by the distance passed 
through by the line; which distance may always be regarded 
as the distance moved by the centre of the line. 

10. If the point A remains fixed while the line revolves 
around it till it, comes again into the position AB from whence 
it started, the surface generated is a circle ; and the line des- 
cribed by the point B, will be the bircutnfer- B 
ence of the circle. If it moves only from the 
position AB to AD, the surface will be a sec- 
tor of a circle, and the line described by the 
point B, is the arc BC of the circumference, 
and because the circumference CEFC des- 
eribed by the centre C of the revolving line 
represents the whole motion of the line AB 
in its revolution; hence, AB drawn into the circumference 
CEFC, represents the whole surface generated. 

11. If the line AB be conceived to de- 8 
crease uniformly, as it is moved forward, 
always in a parallel position, till it termi-. 
nates in a point, the figure generated by the 
motion of the decreasing line AB, will be 
a triangle ABC. 


We may, instead of supposing the tri- 
angle to be generated by the line AB, de- 
creasing as it advances toward C, sup- 
pose it to generated by an infinite series 
of decreasing ordinates, parallel to AB; Ava ce g 
each of which may be supposed to be drawn sanaaeh the ic? 
tance, between itself and the next one. Thus, let ad, ef, &c., 
be a series of decreasing ordinates situated equidistant from 
each other on the line AC, and let ab, be drawn into the 
position Al on the line AB, cd, into the position al, ef, into the 


148 PRODUCTION AND RESOLUTION OF 


position cl, &c., through the whole series. Now, since at the 
extremity of each of the consecutive ordinates a triangle B1d, 
bld, &c., is left without the triangle ABC, which has not been 
described or passed over by the ordinates, the triangle is there- 
fore not perfectly described by the decreasing ordinates; but 
if the number of the ordinates are infinitely increased, those 
spaces become indefinitely small, and hence may be omitted 
as being of no appreciable value. Moreover, if the motion of 
the ordinates is reversed, or if they are moved from A toward 
C, taking AB as the first ordinate. the surface thus described 
would exceed the triangle itself, in the same amountas it would 
fall short in the former case; thus, let AB be brought in the 
position a2, ab, into the position c2, &c., and the small trian- 
gles formed above the hypothenuse BC, by this means, would 
be equal to those falling below it in the former case, and half 
their sum would be a correction to be added or subtracted in 
either case; but since this correction is equivalent to the sur- 
face generated by half the line AB drawn through the distance 
Aa, it follows that the sum of all the ordinates ad, ac, &c., 
+half the line AB drawn into the common distance Aa, gene- 
rates an area equivalent to the triangle ABC. 

«x But when the number of ordinates are indefinitely increased 
one-half, AB is infinitely small in regard to their sum, and 
hence may be omitted. 

12. Hence, if an infinite series of equidistant and parallel or- 
dinates to a right line AD, decrease uniformly from AB to 
AC, then will the surface generated by drawing those ordi- 
nates through or into their common distance, be a trapezium. 

13. If a surface be drawn through any E 
space not in the direction of a line parallel 
to the surface, the product will be a solid. 

Thus, if a surface ABFD be drawn 
through the distance DG, till it comes into _ fii 
the position IEGC, it will thereby generate AX} 
a solid AC. 

If the surface is always parallel to its first v = 
position, and if any point A, moves through a right line AI, 
the solid will be a prism, which will also assume a character 
according to the figure of the generating surface: thus, a rect- 
angular, a triangular, or a polygonal prism, may be generated 
by the motion of a rectangle, a triangle, or a polygon; and in 
the same manner may a circular prism, or cylinder, be gene- 
rated by the motion of a circle, always parallel to its first po- 
sition, and in a direction perpendicular to such generating 
surface, ' | 


GEOMETRICAL MAGNITUDES. 149 


If instead of one surface being drawn through the whole 
distance AB, there be a series of similar surfaces parallel to 
each other, and if each be drawn through their respective dis- 
tances, the same solid would thereby be generated. 

The same principles may also apply to solids of revolution 
of any figure about a fixed axis, and the partial revolution of a 
series of similar figures, one of whose several sides is the com- 
mon axis. 

14. If a plane quadrilateral surface ABGF, E H 
be conceived to move uniformly along in a Rew gett 
direction perpendicular to itself, and to de- 
crease uniformly in one of its dimensions 
during its motion, till it terminates in a line; 8 
it will, by that means, generate a wedge or ‘ 

a triangular prism, ABEF HD. 

It. instead of the decreasing plane, there be an infinite series 
of quadrilateral planes equidistant from each other and de- 
creasing in one of their dimensions in consecutive order, and 
in arithmetical progression till one of them terminates in a line, 
then the series of planes drawn through their infinitely small 
distance will generate a wedge or a triangular prism; all of 
which becomes evident by reterence to Art. 11, tor the same 
reasoning will apply here as in that case, since this prism may 
be conceived to be generated by the perpendicular motion of 
the triangle, which was there found to be the product of an 
infinite series of decreasing lines. 

15. Hence, if across any plane figures, A, B, C, an infinite 
number of parallel and equidistant ordinates are drawn, the 
sum of the ordinates intercepted by each may be regarded as 
a measure of their surfaces when compared with each other ; 
although lines, however associated, cannot represent surface in 
absolute terms, yet an infinite number of parallel lines drawn 
equidistant across two or more surfaces, will represent the ra- 
tious of those surfaces to each other, and may hence represent 
those surfaces in relation to their forms and comparative mag- 
nitudes. 

Hence, for the purposes of investigation, the properties of 
geometrical magnitudes and their relations, an infinite series 
of parallel ordinates drawn across any plane figure, may be 
regarded as the measure of that figure. 

And also, for the same reasons, may an infinite series of pa- 
rallei and equidistant planes passed through a solid, represent 
the capacity or value of the solid. 

Neither will this mode of investigation lead to any error, 
seeing, we do not thereby establish any absolute measure for 
the surface or solidity, in terms of superficial and solid units, 


150 PRODUCTION AND RESOLUTION OF 


but only the relations of certain surfaces or solids through this 
medium, to certain other surfaces or solids with which they 
are compared ; and since they are compared with each other 
under the same circumstances, the result of that comparison 
must hence be correct. 

16. Let the rectangular prism AH, whose 
base ABDC is supposed to be a square, be y YW Yu 
divided as at Prop. IV. B. 1, into the pyra- “7 
mids ABDCE, EFHGD, CDGE, BEFD, 
through which let a plane hodc be passed, — 
cutting the several pyramids in the sections 
mes ys iopg, gpdn, parallel to the base | c= = 
A . Let hi or hs=a,andia ordn=b, | fHfff77Gy 
then may the section hode be expressed by La 
a’+2ab+b’, that is, the section higs=a’, 
(sgnc--iopg)=2ab, and gpdn=b*; Va’-+-ab is a mean propor- 
tional between a and a+b. Let an indefinite number of planes 
be passed through the solid parallel to the base, and each sec- 
tion may be expressed in the same manner, but the value of a, 
it will be seen, is constantly decreasing in arithmetical pro- 
gression, as we ascend from the base to the vertex E.; and 
hence, represents successively a series in arithmetical progres- 
sion; the value of b is also increasing in the same order du- 
ring the successive ascent of the series of planes ; buta+bisa 
constant quantity during the whole change of the relative va- 
lues of a and b, and hence a+ represents a successive series 
which is constant. Now, because a’+2ab, the square of the 
proportional mean between a, one of the series of arithmeticals, 
and a+b, one of the series of constants, which repsesents a sec- 
tion hine through the two pyramids ABDCE, CDFE; it fol- 
lows that since an infinite series of parallel sections represent 
the whole of these pyramids, that the sum of the squares of 
the whole series of geometrical means will represent the whole 
of the solid ABDCEG. But it has been shown, (Prop. IV. 
B. I,) that the solid ABDCGE is equal to one-sixth of the pro- 
duct of the sum of the squares of the two bases plus four times 
a middle section drawn into its altitude. Hence, if there be any 
infinite series of quantities, such that the terms are severally 
geometrical means between the several terms of a series of 
arithmeticals, and of a similar series of constant quantities ; 
then will one-sixth of the sum of the square of the first and 
last terms plus four times the square of the middle term drawn 
into the series, be equal to the sum of the squares of the series. 

Hence, if an infinite series of quantities varying in arithmetical 
progression, be drawn into a similar series of constant quantities, 


d 


Nae, 
= or-¢" 


GEOMETRICAL MAGNITUDES. 151 


then will the sum of the series of rectangles be equal to the~ 
product of the first, plus the last term, plus four times the mid- 
dle term drawn into one-sixth of the series. 

17. It will appear that since the solid ABDCGE represents the 
sum of all the, a?+-ab, viz., the squares of the whole series of 
mean proportionals between the corresponding terms of two 
other series; if there be a series of quantities, decreasing 
from z to 0 in arithmetical progression, and another equal se- 
ries of z a constant quantity, the sum of the squares of a series 
consisting of mean proportionals between the corresponding 
terms of the two series, will be equal to half the sum of the 
squares of the series of z, or equal to half the square of z drawn 
into the series. 

18. It will also appear, that if there be an infinite series of 
quantities decreasing from z to p in arithmetical progression, and 
another similar series of constant quantities, z, the sum of the 
squares of a series of mean proportionals between the cor- 
responding terms of the two former series, will be equal to 
half the sum of the squares of the series of z, plus one-half the 
sum of a similar series of rectangles of p Xz. 

For, let AE be a prism, the length AB E 
or AC, of whose side is equal z, and if 
we make IG and LO=p, and construct 
the plane LIBD, we shall have the pris- 
moid ABDOCGIJL, which may be repre- 
sented by the sum of the squares of a se- 
ries of geometrical means between the 
terms of the series of the decreasing arith- 
meticals, and those of the constant quan- 
ties. But this prismoid may: be further 
divided by the plane GOCB into the wedge ABCDOG, which 
is equal to half the sum of the squares of the series of z, and 
the wedge GOLIBC, which is equal to half of a similar series 
of the rectangles of z into p. 

19, Let a series of s, be the series of constant quantities, (and 
let s=AD,) while another series varies from z to p in arithme- 
tical progression, (making z=AB, and p=GJ.) Then will the 
sum of the squares of the series of geometrical means. between 
the terms of the two former series be equal to half the sum of 
a similar series of rectangles of sXz+a series of rectangles 
of sXp. | 

For the prismoid ABCDOGIL equals the series of mean pro- 
portionals as before, and the wedge ABDCOG will be equal to 
half a similar series of rectangles of the series s with a similar 
series of z, and the wedge IGHLCB will be equal to half the 
sum of a similar series of s, drawn into an equal series of p. 


152 PRODUCTION AND RESOLUTION OF 


Hence, a wedge may be conceived to be generated from a 
series of squares of mean proportionals between a series of 
constant quantities, s or z, and an infinite series of lines in 
arithmetical progression from z to 0, or from o to 2. 

20. Let AD, ad. ad &c., be an _ infinite 
series of lines, increasing in arithmetical pro- 
gression from 0 to z, or from E to BA, and 
the arithmetricals will generate or constitute 
the triangle ABE, and the sum of their squares i\ 
will generate or describe the pyramid ABCDE. , =F = He 
For if an infinite number of parallel lines or Sr ERNIE, 
ordinates be drawn across a plane figure, 4 
those ordinates will represent the figure in the relation of its 
magnitude and form; (Art. 12,) but the lines or ordinates, 
ad, ad, are infinite in number by hypothesis, and increase from 
0 to z, or they commence at EK and terminate at AB; hence 
the sum of the ordinates constitute, or are a function of the 
triangle. Also if an infinite number of parallel planes be 
passed through a solid, the sum of those planes drawn into 
their distance, may be regarded as the solid when compared ; 
for the solid consists of an infinite number of parallel planes 
drawn into their infinitely small distance ; (Art. 16) and hence 
this series of planes would represent the solid in the re- 
lation of its magnitude, as the ordinates represent the tri- 
angle. _ 

But if CB = AB, then each of the parallel planes, adc, adc, 
&c., will be the square of its corresponding ordinate, ad, and 
all the planes will be the sum of the squares of all the ordi- 
nates. Hence, the sum of the squares of the series represents 
the pyramid, ABCDE, in the same manner as the sum of the 
ordinates represents the triangle, ABE. 

21. If the serics of lines be a constant 
series of z, then in the same manner as the 
series of arithmetricals would describe or 
generate the triangle ABE, the constant 
series of z would generate the circumscrib- 
ing rectangle ABFE, and the series of 
squares of z, would generate in like man- 
ner the prism ABCDEFHG, circumscrib- 
ing the pyramid. ‘ 


A 
If instead of the squares of the several terms of the series, 
in the proposition generating the pyramid, there be taken the 
series of circles described on the several terms as diameters, 
those circles would generate a cone = the cone inscribed in 
the pyramid. And if instead of the squares constituting the 
prism, Art. 1, there be taken a similar series of circles, then 


GEOMETRICAL MAGNITUDES. 153 


would the solid generated be a cylinder equal to a cylinder in- 
scribed in the prism. Also, if we take any other figure, which 
we constitute a similar function of each successive term of the 
series, We may generate pyramids, prisms, or frusta, of differ- 
ent character, but of the same general species. 

21. If a line be made to increase from p to z, and if its va- 
riation be represented by an infinite series of arithmeticals, 
then the series will truly represent a trapezium, and the series 
of the squares of the first, may represent a frustum of a py- 
ramid. 

' 22. If a be made to pass successively through all the values 
from 0 to z, while 6 is made to pass in like manner through 
all the values from U to p, then the two series drawn into each 
other will generate a pyramid with a rectangular base. 

23. If there are two variable lines a and 
6, and if a be made to pass in succession 
from p to z, or from z to p, inan infinite series, 
while 4 remains equal to z or s, representing 
a similar series of constant quantities, then 
will the solid produced by drawing the cor- 
responding terms of these series into each 
other represent a prismoid AG; but in this 
case the prismoid will have two parallel 
sides ABIE, CDFG. But if, while a@ is 
passing from z to p, 6 at the same time varies successively 
from z or s to f, then will the solid generated by the series of 
rectangles of the corresponding terms of the variable quanti- 
ties, be a prismoid, neither of whose sides would be parallel 
except the two bases. 

24. If one of the variable magnitudes should be made to 
pass successively in an infinite series from 0 to z, while the 
other should pass from z or s to p, then the solid generated by 
the rectangle of the corresponding terms of the series would 
be a wedge. 

If a be an infinite series of lines, vary- 
ing from 0 to z in arithmetrical progres- 
sion, and / be a like series varying from 
s or z to 0, then if the corresponding terms 
of the series be drawn into each other, 
their product will be a triangular pyra- 
mid which may be resolved into the two 
wedges sengCD, and scngEG; all of 
which is evident by inspection. ‘ aq: 

Let the ordinates drawn across the triangle CD, parallel 


it 


> 
bo 


11 


154 PRODUCTION AND RESOLUTION OF 


ordinates drawn across the triangle EGD, be a series decreas- 
ing in the same time from IG or s to 0; call this series b: 
then the series represented by a. drawn into that represented 
by b, will generate the solid EGCD. 

25. Let the ordinates drawn across the triangle ABC be a 
series decreasing in arithmetical progression from z, or AB 
to 0, and let those drawn across the triangle BCD bea sim- 
ilar series of ordinates increasing from 0 to z, then may 
the series of proportional means between the corresponding 
terms, represent a series of equidistant ordinates, drawn 
acrossa semi-circle, whose diameteris z= DB, and the rectangle 
of the corresponding terms. may represent a similar series 
of equidistant ordinates, drawn across such portion of a 
parabola whose axis is AB, as is intercepted by the curve, 
and a diagonal from the vertex B to the extremity of the 
base, the axis of the parabola being equal to <. 


For the ordinates dn drawn across a ¢ 
semi-circle are severally mean propor-  —=====4>=> 
tionals, between the abscissze of the 
diameter; that is, any ordinate dn, isa 
mean proportional between dB and 
dD; now if their be a series of ab- 
scisse dB taken in arithmetical pro- 
gression increasing, then their cor- 
responding abscisse dD, dD, &c.,, B 
will be a series of decreasing arithmeticals ; and if BD is 
equal toCD or AB, then will the abscissee dD, dB be severally 
= to their corresponding ordinates de, ef, in whatever position 
they are taken ; hence the series of ordinates, across the semi- 
circle is a series of proportional means between the several 
corresponding terms of the increasing and decreasing series. 

Again, it has been shown (Prop. VII, Sch. B. I,) that the 
expression for any ordinate ec, drawn across the parabola, 
CBh, between the curve and the diagonal, parallel to the axis, 
is equivalent to the rectangle de X ef, and since this is true of 
every parallel position of the ordinate ec ; hence the sum of 
the series of ordinates, is equivalent to the sum of the series 
of rectangles of the series of variables, = the sum of the se- 
ries of squares, of a series of ordinates dn, &c., across the 
semi-circle. 

Let DB be greater than CD, or AB and the ordinates dn, 
will be equivalent to those drawn across a semi-ellipse, whose 
major axis is BD, and minor axis CD; but if CD is greater 
than BD, then will BD be the minor axis of the ellipse, &c. 

Hence the solid ABCD formed by drawing the correspond- 
ing terms of the increasing and decreasing lines into each 


| 


GEOMETRICAL MAGNITUDES. 155 


other is equivalent in its expression to the area CBh of the 
parabola, and the semi-circle DBn is equivalent to.the sum of 
a series of square roots of an infinite series of parallel or- 
dinates ec across the parabola, orof parallel sections defi through 
the solid. 

26. The elements contained in the forego- 
ing propositions, may be applied also to 
solids of revolution ; for instead of a prism 
and its inscribed pyramids in Art. 16, we 
may substitute a cylinder, and its inscrib- 
ed cones, &c. ; and if planes HPNQ be 
passed through the cylinder, the cones 
and their complements, these will be cut 
in the relation of their magnitudes, respectively. in each section; 
which will be in all cases, the same as that of the prism, and 
the inscribed pyramids, of equal base and altitude. 

If the semi-circle or semi-ellipse CIN be made to revolve 
about the axis CI, and by its revolution to produce a sphere, 
or spheroid, then, because every section described by the pa- 
rallel ordinate ON, would be proportional to the squares of their 
circumscribing lines respectively, the sum of the sections de- 
scribed, would be proportional to the sum of the squares of 
the generating lines ; hence the sum ofan infinite series of those 
sections, and consequently the solid generated by the semi- 
circle would be equal to four times a middle section, multiplied 
by 1 of the series. And any segment or zone of the sphere 
or spheroid, will be equal to the sum of the bases + four times 
a middle section X } its altitude. 

27. If every section defithrough the pyramid ABCB (Art. 25) 
should be contracted in one of its dimensions tillit becomes a 
square, and if the edge DB should continue to be a right line, 
then will the sides BCD, ABD, be plane surfaces, and the 
side ACD, ABC will become curved, and the whole solid will 
be equal and similar to a quadrant of a revoloid ; and four of 
such pyramids, would constitute a perfect right revoloid ; 
moreover the sides ABC, and ACD would become semi-cir- 
cles. : 

Forit has been shown that the square 
root of every section defi, through the solid, 
is equal to an ordinate drawn across the 
semi-circle DnB through the same _pa- 
rallel ; but the square root of the section, 
is the side of a square equivalent to the 
section; hence, if the sides ef the sections 
are severally the ordinates belonging to a | 


156 PRODUCTION AND RESOLUTION OF 


semi-circle, then the solid must acquire a form, similar to 
that of a quadrant of a rectangular revoloid, such as would be 
formed by passing planes through the axis, bisecting its oppo- 
sides. ' 

If, when the sections of the solid become squares, its vertices 
are conceived to be at the extremities of an axis, passing 
through its centre ; then the solid would become an elliptical 
revoloid ; its conjugate axis being = to } its vertical or trans- 
verse axis. 

The pyramid ABCD is equal to } of the prism, whose base 
is the square of AB, and altitude BD ; four such solids, or the 
whole right revoloid is = 4 = 2 the circumscribing prism. 
Also, the prism, circumscribing the elliptical revoloid, whose 
base = defi, and altitude BD, being = 1 of the former prism, 
is the prism circumscribing the elliptical revoloid ; hence, the 
elliptical revoloid is = % its circumscribing prism. 

If from a cylinder of equal base and altitude, two equal 
cones be taken, one on either base, and of an altitude equal to 
that of the cylinder, the two remaining portions would be each 
equivalent to the cylinder; and every section through each of 
these portions, by planes parallel to the cylinder’s base, would 
be equivalent to a corresponding section through the quad- 
rant of the sphere ; each of these portions are = 2 of the cyl- 
inder ; hence, four of these portions = the sphere, are = 2 the 
circumscribed prism as found in the Elements of Geometry. 


Scholium. We may, from the preceding investigations, 
draw the following deductions and conclusions. 

First, that any series of quantities in arithmetical pro- 
gression, varying from z to 0, drawn into any series of 
constant quantities, will produce a quantity whose value is = 3 
the sum of the base or maximum product of the variable + 
four times the value of the product of half the maximum value 
drawn into the series. Also, that the value of any multiple of 
this series may in like manner be determined. 

Second, that if any series of numbers varying from 2 to 0, 
in arithmetical progression be drawn into another similar 
series, direct or reciprocal, the value of the product is = 3 
the sum of the two bases produced + four times a middle base 
formed by drawing 4} the maximum values of the terms of the 
series into each other. 

And that this is true for any multiple, or power of the va- 
riable series, whose exponent is an integer or when, any 
number of variable quantities are drawn into each other, whe- 
ther direct or reciprocal, and this is the basis of the Integeral 
Calculus, as will appear in the subsequent pages. 


GEOMETRICAL MAGNITUDES. 157 


CHAPTER ITI. 


¥. 


ON THE CONSTRTCTION OF QUANTITIES WHOSE ELEMENTS ARE 
A SERIES OF CONSTANT OR VARIABLE QUANTITIES. 


Art. 1. Having proceeded thus far, in analyzing the pro- 
ductien of geometrical magnitudes, showing the manner and 
Jaw of their generation, we are enabled, by having the ele- 
ments and the law of the production of any magnitudes, to 
give a geometrical construction of such magnitudes. 

We were taught, in the application of algebra to geometry, 
the mode of constructing integral algebraic quantities or ex- 
pressions geometrically ; we are now to represent a series of 
quantities, under asingle construction; or to construct quanti- 
ties whose elements are a series, either of constant, or variable 
quantities. | 

2. In considering the relations which exist between different 
quantities, those which, during the whole of any investigation 
are supposed to retain the same value, are called constant 
quantities ; those to which different values are assigned, are 
called variable quantities : constant quantities are usually re- 
presented by the former letters of the alphabet, as a, b,c, &c., 
and variable quantities by the latter, as uxyz, &c. 

3. When two or more variable quantities are connected in 
such a manner, that the value of one of them is determined 
by the value assigned to the other, the former is said to be a 
function of the other variables. 

Thus, in the equation y=azr+bz’?+c, where the value of y 
depends on the value assigned to 2; y is said to be a function 
of x, which is usually expressed by f’(z), 9(x), (2), or similar 
abbreviations. 

Also, if an infinite series of equidistant ordinates are drawn 
across a surface, the sum of those ordinates is a function of 
the surface. So, also, the sum of an infinite series of planes 
through a solid may be regarded as a function of the solid, or 
the solid or surface a function of the planes or ordinates. 

4, When any quantity or magnitude as the element of other 
quantities or magnitudes is variable, the sum of a series of the 
variable quantity, within its variable limits, may be represented 
by a dash drawn below or above the letter representing the 
variable quantity. 

Thus z or 2 may represent a series of the variable quantity, 
z. If the incipient value of the variable is z, and the series is 
decreasing to s, or 0, the dash must be placed below the let- 


158 PRODUCTION AND RESOLUTION OF 


ter ; but if its incipient value is 0 or s, and its terminate value 
is z, being an increasing series, the dash must be placed above. 
Thus z iudicates a series decreasing from z to s or 0; and Zz 
indicates the series increasing from 0 or s to z 

5. If it is required to express a series of quantities in arith- 
metical progression from z to s, or from s to 2, it may be thus 
written—z2: -- -s, or Z++s3 or which is the same,@hst! OT Lesa, 
x being the incipient and 2’ the terminate value of the series. 

‘The condition of questions involving these variables, gene- 
rally indicate the incipient and terminate values of the increas- 
ing or decreasing variables. 

6. The expression z is equivalent to that of Z, when consi- 
dered independent of other variables; but the product arising: 
from drawing z into Z is not equivalent to that ‘of drawing z 
into z; for zz indicates that the greatest value of z is drawn 
into its least value, and consecutively ; and zz indicates that 
the greatest value of z is drawn into the greatest, and so on 
through the series. 

A series of the squares of z or 2, is represented by z’ or 2? ; 
a series of roots by /z or V2. 

7. When it is designed to express, a series either of con- 
stant or variable quantities, without regard’to their progres- 
sion or law of variation, a small capital, of the letter denoting 
a single term of the series may be used. 

Thus in the equation y= /dz, if we would express a series 
of y, it may be written y, and the equation will be y= Ydz 5 
this, unless otherwise restricted, expresses an infinite series of 
the quantity represented by y. If y= v(dz) is the equation: 
of any figure, then y, is a function of the surface, or y= /(dz)x 
is the equation to the surface. 

Hence the equation to a surface consists of the equation of 
the figure considered as a series, drawn into the axis or ab- 
scissa. 

8. If a be the magnitude AD of any series whose number 
is n, of lines AB, and whose length is the constant quantity z, 
then the sum of the lines will be nz; and their magnitude 


a 
made by drawing them into a surface, will be qth a 


| i: 
a 


Tor we have shown that if an infinite series 
of lines are drawn into their respective dis- 
tance, the product is a surface ; hence, we have 
the following construction, viz: a rectangle. 


ABDC. a 


GEOMETRICAL MAGNITUDES. ~ 159 

9. Let a be a series of lines z, con- 
stantly decreasing in arithmetrical pro- 
gression from z to 0, or from the line AB 
to the point D, while the line BD repre- 
sents the number or magnitude of the se- 
ries; and the series of z drawn into the 
quantity a, will be equivalent to the tri- 
angle ABD. 

If a be a series of lines, AB decreasing from z to s, or from 
AB to EC, then if AD=a = the magnitude of the series, the 
construction will be the trapezium ABCK. 

10. If a be a series of z*, where z decreases uniformly from 
AB, to a point D; or from z to 0, then az? may be constructed 
by the exterior space of a parabola AeDB, for while az ge- 
nerates the triangle ABD, a.z’ will generate the parabola 
AeDBA, (Prop. VII, Sch., B. 1,) whose axis is CD. 

C 


D 
If a be a series of Vz, then may az be 

put under the construction of a semi-parabola, 

AcDB, whose axis is DB=a, and whose base 

AB=z, the series of z being a series of ordi- 

nates across the triangle ABD, parallel to AB. 


A B 

12. Let it be required to construct a quantity dV (z.z), or a se- 

ries of mean proportionals between the corresponding terms of 

two equal increasing and decreasing series drawn into d, the 
number of the series. 

If d = the line AB, and if z = the same line, the surface 

generated by drawing d into /(z.z), would be the semicircle 


ABD. 


D 
For the equation to the circle is y’= / (dx—z’) 
Gr me DC af AGL CL. 


AC B 

But if z be greater or less than d, then the construction will 
be a semi-ellipse ; which if z is less than d, will have AB for 
its major axis; butif z is greater than d, AB will be the minor 
axis. 

For the ellipse by its equation, d?: ¢:: 2(d—z) :.y’ or d*y?’ 
=c’(dx — x’) is only the circle expanded, or contracted, in the 
ratio of the major and minor axes. 

The equation for the surface of a circle or an ellipse, may 
more properly be expressed by d./(xx) for a semi-circle or 
semi-ellipse, and if x = d, then the equation will be that of a 


160 PRODUCTION AND RESOLUTION OF 


circle; but if d and x are unequal, the equation becomes that 
of an ellipse, and d will be the major or minor axis, according 
as it is greater or less than 2, and 2 will be its conjugate ; and 
its equivalent in either case is 1}dxX+, or for the whole circle 


or ellipse 1dzxX: hence 7 isa function of ./ (xz) or fv (Er) 


Hence we have a finite expression for the circles quadrature 
in algebraic terms, viz: 2d./(x.) or 2x /(z.2)=r'x, where x 
is equal to the diameter. Andr=(2r+r") / (z.%)=the circum- 
ference of any circle whose diameter is x ; z and x being series 
of increasing and decreasing quantities. Hence, t=2d/ (xx) 
=1d'=(8~d) y (a2), or (82) V (22). 

13. Let ABD be the segment of a 
circle, the height of the segment DE 
being equal z; and if the diameter 
= 2, then will the chord AB= (zz 5) 
and if EG = x—z = 2’, then will the 
equation to the surface be 2yz=2z 
Bice 2)2). 4 ach erpmnvenlte: minh 64) 


34 

Also, let CE = 1x—z=u, then if AB in a decreasing series 
is drawn into CH, or if (u./(xz)) be constructed, it will be equal 
to the triangle ABC ; and the sum of the segment and tri- 
angle, = 2z/((z-:x')z) + u(/(xz)) = the sector ACBD. (2) 


In the triangle ABF the side AF is equal 2CK=xz—2z ; and 
BF:BA:: AF: AS, or 2: xz: :a2—-2z: AS the sine of the 
angle, BCA, or sine of the arc of the segment 

=f (zr) — 2zV (zz) +r. ee ee ee (8) 

Let AS=s, and let the are ADB=7’ and the area of the 
segment, (Prop. XVIII B. IV) will be (#’—s) xX{z, hence the 
expression 22./((@-- 2’)z)=1a0' — jars: - - - + = (4) 

Therefore the arc ADB of the segment may be expressed 

2 af ((2-°2')z)-+s % Jf (LX 
of = VIE ETS = ary (eat VC) 6) 
4 

14. If a series of variables consisting of two or more fac- 
tors, as its elements, are drawn into the number denoting the 
series, the construction can more conveniently be represented 
by a solid. 

Surfaces properly consist only of factors, equivalent to the 
second power ; cubes of factors equivalent to the third power, 
or the product of three factors. a 

But we may, under certain conditions, construct quantities 
representing solids as surfaces ; those whose factors are con- 
stant, by any surface arbitrarily chosen, and those which are 
variable by such curves as yield to the conditions of the ex- 
pression ; and we may thence proceed to construct such solids 
as would result from drawing such surfaces into a series ; or 


GEOMETRICAL MAGNITUDES. ~ 161 


such as would be the result of drawing the variable series of 
elements of such surface into a constant, or a variable quan- 
tity ; and the quantity so constructed may, by being put under 
a superficial construction, become the base of another quantity 
or magnitude, made by drawing this quantity into another 
given quantity or into a series; and we may proceed, in this 
manner, to construct quantities geometrically, converting one 
construction into a base for the next, and so on. 

This may be performed algebraically, and with greater fa- 
cility, inasmuch as a quantity involving any power of the va- 
riable may be assumed as a base for a higher power of the 
series, and thus the powers and roots of variables may be 
extended at pleasure; and the laws of variation or their in- 
crease or decrease in value, may be determined by geometri- 
cal construction and analytical deductions. 

15. If it were required to construct a series, a, of the quantity 
bc, drawn into each other, b and c being constant quantities ; 
here it is evident that, for the purposes of construction, be may 
represent either a line or a surface; if the series of be be re- 
presented by a line, then bc, drawn into a, or abc, will re- 
present a rectangle. But, if bc represents a surface, then abc 
will represent a prism. And for the purposes of investigation, 
the conditions of the quantities would be similar in either case; 
for the line would have the same relation to the rectangle ge- 
nerated by drawing the line into the series, as the surface to 
the prism, generated by drawing the surface into the series. 

16. Let it be required to construct a series a, of bz or a.bz, 
where z is a series decreasing uniformly to o. 


This would evidently be represented under 
the form of a wedge ABCDEF, whose base 
ABCD is represented by bz, and the perpendi- ,, 
cular ED represents the value of a. 


A 

If z decrease only to s, then the construction would be the 
prismoid ABCD abcd, whose greater base, ABCD, is repre- 
sented by 6z, and whose lesser base, abcd, is represented 
by bs. 

17. If « be also a quantity 
decreasing uniformly with z, 
in such ratio as to reach o at 
the same time, then the ex- 
pression a.uz would generate 
a pyramid ABCJ; which 
will be its proper construc- 
tion; and if w=z, or if the . 
expression can be but under | 
the form a.z’*, then the pyramid will have a square base. 


162 PRODUCTION AND RESOLUTION OF 


18. Let a series a.zz, be constructed. This quantity, Sheet 
senting the products” of the corresponding 
terms of two series, one increasing from o to 
z, while the other decreases from z to 0, drawn 
into a, the quantity denoting the number of the 
series, generates the double wedge ABGD, 
where AB, or GD, is equal to z, and the per- 
pendicular AD=a. 

If uw represent a series of ordinates drawn 
across the triangle ADG, then az will re- 
present the triangle; and if these ordinates are severally drawn 
into a series of uw, the result would be the same as before. 

19. Let it be required to construct y°=pz, which is the 
equation to the vertical parabolic revoloid, y= y(pz,) being the 
equation to the parabola, p being the parameter, is a constant 
quantity, we have z variable. 

First, we may draw p into a series of 2, which gives us the 
triangle DEA (see diagram to art. 16,)=pz, p being equal to 
DA, x=the axis of the revoloid=DE the altitude ; Satie if this 
triangle, as a base, is also drawn into z, the axis of the revo- 
loid, we shall have the solid ABCDEF. 

Or, more properly, if we first draw the axis z into the se- 
ries of the variables, we shall have the triangle as before, with 
a base AD; and if we multiply this by p=DC, we shall have 
the solid described as before, then will any section abed of this 
solid parallel to the base ABCD, be equivalent to a similar 
section through a parabolic revoloid, or pyramoid, which it is 
designed to represent. 

The expression for a cube may at all times be constructed 
on a surface by means of curves; but every different species 
of solid requires some peculiar construction, according to the 
equations for the ordinates or sections of the solid, the expres- 
sions for which may be transferred to expressions for ordi- 
nates to a surface, or equations to some curve; and since any 
multiple, or power of a series x, whose exponent Is an integer, is 
known, when the series or root is known; we can hence dis- 
cover an innumerable variety of curves, which are quadrable ; 
but, in general, in descending powers, it is not certain that the 
series of roots of a given series may be so. Thus, the circle is 
equivalent to a series of square roots, of the ordinates to a pa- 
rabola; the series of the parabala is quadrable, but not the 
circle, except in certain functions of given quantities; we may 
get an expression for its value, but it will be under an incom- 
mensurable form. 

Let ABD represent the vertical segment of a rectangular 
spherical revoloid ; and its equation considered in relatinn to 
its figure, will be y* = (z.. 2’)z - - - - (1) 


GEOMETRICAL MAGNITUDES. 163 


the equation considered as a solid, will y°z = z(z.. a')z (2) 
which from its organization will readily be discovered to 
be cubable, for its value is a (x'z + 4 (Lx + 42’) 42). = 222? 
nA RMR i 2s ty ing tths esha ®) 
itsconvex surface will be 4xz - thawen Shes pe (4) 
Hence, the solidity of a spherical segment will be = 
(2227+ 42'2%)le = Lazo + la'2e - - - - (5) 
and its surface will be xz - - - . : (6) 


2 


; 7 c ; 
20. Let it be required to construct v'=7,(dr¢ +2") the equation 


to an hyperbolic revoloid, vy’ being a series of parallel planes, 
being the squares of the ordinates to the axis. 
Because this equation involves powers higher than the 


Cc 


square ; that is, because the terms (dz+z’,) multiplyed by 5 = 


produces a solid, the series represented by these ane 
drawn into a, would be of a higher power than a cube ; hence, 
in order that our construction shall be under a cubie form, @ 
(dx +2”) must be represented onaplane. Let a=CB. 
Hence, we have a.z*=the exterior pa- G 
rabolic surface ABC, AB being the axis 
of the parabola, and ufo dé = CBE; 


now, if this is drawn into 7 we shall have 


the solid ABECFDG; for the construc- 
tion of a solid equivalent to the quadrant p 
of an hyperbolic pyramoid or revoloid ; 
every section NIKL through this solid 
is equivalent to a corresponding section through a hyperbo- ~ 
lic pyramoid of the same altitude and equivalent base. 

Hence, it will be perceived that the hyperbolic revoloid is 
cubable; for this, its representative is equal to one-sixth of the 
product of the sum of the base AEFD plus four times a mid- 
dle section NIKL drawn into the altitude BC. This-is also 
true in relation to the parabolic revoloid: 

And since a paraboloid or hyperboloid is to its respective 
circumscribed revoloid as a circle to its circumscribed square, 
or as r°x : 4r’, it follows that the paraboloid and hyperboloid, 
are also cubable in terms of =. 

21. It may be perceived that in constructing the quantity az, 
where x is a series decreasing to 0, we have a triangle ; hence. 
the value of az is tax; or the value of ex may, in like manner? 
be shown to be ryt, ’ 

Also, since 2°a is equivalent to a pyramid whose base is 2”, 
and whose altitude is x; henee, z°x is equal to 32°. 


Ss 


164 PRODUCTION AND RESOLUTION OF 


And if we have a series of z* to be drawn into 2 or a, we 
may also construct 2° by art. 20, and we shall here have 2°z 


=127'; alsoz*z= “ihas My 

Hence, the law of the progression of the powers of the va- 
riable quantities is manifest, for the value of the continued pro- 
duct of any series of power, of a series of variable quantities 
drawn into any constant quantity, is equal to the same power 
of the maximum value of the variable drawn into the same 
quantity, divided by a number denoted by the index of the 
power plus 1. 

If the variable has a fractional exponent, the same consi- 


1 
Moy 


deration will also apply. Thus, a xis equivalent to 3a and 
yee 


4 
x*x is equivalent to 2z m} anda" x equivalent to cana &c. 

Where the product of any power of a variable quantity is 
made the base of another construction, according to its deter- 
minate value, the magnitude constructed on this base, is no 
longer considered as a function of the variables, entering into 
the former constructions as elements of this base; but this 
new construction is subject to the laws of its own organiza- 
tion; and if raised to any power, or multiplied by any number 
of factors, either as constants or variables, the different powers 
depend on, or are a function of the base assumed, and not of 
its original constituents. 

Thus, if we have the quantity 2°z, its value is }z*; now, 
if 1z* is assumed as the base of another construction, as (12“)z, 
or 1z‘z the product is not =;2°, but is =1z°. 

Ifa series uxz are drawn into a, the product is equivalent 
to }uzza; the co-fficient being the same as that for the pro- 
duct of z*z, which gives }2* as its equivalent. 

When the terms of the variable series denoted by x or z, 
are drawn into each other according to their reciprocal value, 
or where an increasing series is drawn into a decreasing se- 
ries, the value of some power of the product of the series may 
be obtained ; and the other powers and roots of this value are 
subject to peculiar laws for theirdevelopment. For the contin- 
ued products of these series, are not subject to the same ratio, 
or the same order in the variation in their successive changes 
from one power to another, as is observed in the products 
of such as are either all increasing or all decreasing. ‘Thus, 
if we have eV 22, let this quantity be involved till its radical 
is removed, and we have z°zz, the value of which may be de- 
termined by a construction. Thus, x%.x2, which is equivalent 
to a double wedge, (Art. 24, Chap. I,) or which will be more 
available for our present purpose, it may be put under the 


GEOMETRICAL MAGNITUDES, 165 


construction of the parabola cut off by a diagonal, (Art. 25, 
Chap. 1,) the value of which is evidently equal 12°x4x1a 
=12°, which is equal to one-half the value of G 

xx; andif this value is again put into a series, "i < | 

and drawn into z, we shall have 42°.2, or \ 
x’z.x, whose proper geometrical construction 
will be the solid CELI, which is the solid of a 
prism circumscribing a parabolic complemen- 
tal ungula, which (Prop. X VIII, Cor. 2, B. II.) 
is equal to one-fifth of the prism CANELI 


—17* 
Tow 


Let there be a series of 2’z, a series of 
xx, and two series zz, and the sum of the 
series will be equivalent to 2°, as may be 
represented by the prism AH. 

For 2x equals the pyramid ABCDE, 
xx equals the pyramid EFHGD, and 2¢%zrzx 


4) 
= 


equals two wedges CDEG completing the : cr ae Tj D 
prism whose value is 2’. 4 Wj YY 
A B 


Let there be a series a of z drawn into a reciprocating Se- 
ries of Z, the terminate value of z being z’, and the incipient value 
of x being 0. Since the product of these series a.(z ++ 2')% 
gives us for a construction a solid hinoCD, whose value is 
equal to jax(2z'+z); and this also represents a segment of 
a quadrant of a revoloid where z and z increase and decrease 
in the same arithmetical ratio, equal to a segment of a cylin- 
drical ungula. 

Also, if we have a series of the products 
of reciprocating variables, or zz drawn in- 
to a series of A or z, its value has been 
found equal to tAz* or 32°, let this be drawn 
into another series of Z, and we have the 
following construction, which is a parabo- 
la whose vertice is D; and we have 222? 
equal the parabolic ungula ADBC equal 
2ABXHAXIBC=2(zX (42X12) X12z)=347'; 
hence, the series zzz’ is equivalent to ;;2°. 

Let this be drawn into another series of Z, and we shall 
have 22z°=12(2?+-4(42° x $z))=12'+.12°; if this is drawn into 
another series of z, we shall. have 22zz*=12z(z°+4(74-2'X12)) 
=}2'+,5;z"; hence the law of progression is manifest. 

If we have a series of zzz’, its value will be 42(2z?+2*) 
=12°+,';2°. Hence, the law of the progession of this series 
will easily be discovered. 


166 DIFFERENTIAL AND 


Scholium. The results obtained by the preceding notations 
and geometrical constructions, are similar to those obtained 
by the integral calculus, the same principles as here used serv- 
ing as the foundation of that science; and in order that a com- 
parison between the two modes of notation may be instituted, 
we will present some of the elementary principles of the cal- 
culus, as the subject of the next chapter. 


CHAPTER III. 
DIFFERENTIAL AND INTEGRAL CALCULUS. 


Art. 1. Asa basis of the deferential calculus, we may premise 
that if y be a function of 2, and if a change takes place in the 
value of f (x) so that x becomes x+h, x being quite indetermi- 
nate, and h any quantity whatever, either positive or negative, 
a corresponding change must take place in the value of y, which 
may then be represented by y’. If the quantity f (ec +h) be 
now developed in a series of the form 

TAL) TAR BiETCh Re... oe 

which is always practicable, in whlch the first term is the ori- 
ginal function f (x), and the other terms ascend regularly by 
positive and integral powers of h, and A, B, C, &c., are inde- 
pendent of h; then the co-efficient of the simple power of h 
in this series is called the first differential co-efficient of y or f 
(x). This is the fundamental definition of the differential cal- 
culus. 

2. Let us now examine the change which takes place in 
the function for any change that may be made in the value of 
the variable on which it depends. 

Let us take, as a first example, 

Uy (ALi, 
and suppose x to be increased by any quantity 2. Designate 
by wu’ the new value which w assumes, under this supposition, 
and we shall have | 
u’ = a(x-+h)’, 
or by developing 
wu! = ax*+2azrh-+-ah’. 

If we subtract the first equation from the last, we shall 

have 

u’ — u=2arh+ah’; 
hence, if the variable x be increased by h, the function will 
be increased by 2axh-+ah’. 

If both members of the last equation be divided by h, we 
shall have 

u'—U 


ici eon 2ax-+ah, (1) 


INTEGRAL CALCULUS. 167 


which expresses the ratio of the increment of the function to 
that of the variable. 

The value of the ratio of the increment of the function to 
that of the variable is composed of two parts, 2axz and ah. 

If now, we suppose / to diminish continually, the value of 
the ratio will approach to that of 2az, to which it will become 
equal when h=0. The part 2az, which is independent of h, 
is therefore the limit of the ratio of the increment of the func- 
tion to that of the variable. The term, limit of the ratio de- 
signates the ratio at the time becomes equal to 0. This ra- 
tio is called the differential co-efficient of u regarded as a func- 
tion of x. 

3. Then let y be a function of z, such that 


y=ax* 
Let x become x+A and y become y’ 
y'=a(x+h)* 
expanding =az>+2az..h-+-ah 


This, it will be perceived, is a series of the required form, 
the first term az’ is the original function y, and the other 
terms ascend by integral and positive powers of h; hence, 
according to our definition, 2az the co-efhicient of the simple 
power of h in this series, is the first differential co-efficient of 
y or f (x) 7 

4. Again, let 


Ue 
Let z become x+h and y become y' 
y'=(xz+h)* 
expanding =2°+32? .h+3z .h?+h* 


Here, therefore, 3z’, the co-efficient of the simple power of 
h is the first defferential co-efficient of x’. 

5. Again, let 

yar’ + ba’ +ca-+td 

Let z become (x+A) and y become y/ 

Re af dia) Ober icc tb) 
expanding = axv*+3axr*h+ 3arh’+h?+b2’+2bxh+-bh?+cx+ch 
+d, arranging according to powers of h, 

=(ax*+bx?+cx+d)4+(8ac?+2bz2+c)h+(Bar+b)h?+h, 
a series of the required form, for the first term is az*+b2’?+cx 
+d, the original function, and the succeeding terms ascend 
regularly by powers of h. 

Hence, 8ax*+2bx-+c the co-efficient of the simple power of 
h in the development of y’ is the first differential co-efficient of 
yf or ax’ ba -+ or + as ‘ 

6. We have now to introduce a notation by which this ratio 
may be expressed. For this purpose we represent by dz the 
last value of h, that is, the value of h which cannot be dimin- 
ished according to the law of change to which h is subjected, 
r 


168 DIFFERENTIAL AND ay 


without becoming 0; and let us also represent by du the cor- 
responding value of «~: we then have 
du 

ett 2az., (2) 

The letter d is used merely as a characteristic, and the ex- 
pressions du, dx, are read, differential of u, differential of x. 

It may be difficult to understand why the value which A as- 
sumes in passing from equation (1) to equation (2), is repre- 
sented by dz in he first member, and made equal to 0 in the 
second. We have represented by dz the /ast value of h, and 
this value forms no appreciable part of h or x. For, if it did, 
it might be diminished without becoming 0, and therefore 
would not be the last value of hk. By designating this last 
value by dz. we preserve a trace of the letter 2, and express 
at the same time the last change which takes place in 4, as it 
becomes equal to 0. 

7. Let us take as a second example, 

u = ax’, 
If we give to z an increment h, we shall have, 
u'=a(a+h)*=ax'+8ahz’+3ah*c+ah*, 


hence, u! —u=8ahz’+3al’xz+ah', 
and the ratio of the increments will be 
u'— uu 
Rap sax’ +3ahz+ah’, 
and the limit of the ratio, or differential co-efficient, 
Ha 3ax* 
dz 7 
In the function 
’ du 
= nz*, we have— = 4nz*. 
dx 


And if y=f (2) 
the first differential co-efficient of y is denoted by the symbol 


d 
oa thus in the above examples. (Arts, 4. & 5.) 


y = ax’ 
hy 

dp ee 
yee 
dy 
de Oe 


y = ax+bz'?+crz+d 
dopant 
se 3ax?+2ba+c 


we 


INTEGRAL CALCULUS. 169 


in like manner if w=f (z) the first differential co-efficient of xy 
or f (z) will be represented by = 


Note.—Since a constant quantity is not susceptible of change, 
it is manifest that it can have no differential co-efficient, or if 


dy 
U feeat 2 pill 0. 


3. To find the first differential co-efficient of any power of a 
simple algebruic quantity. 


Let y=2" 
Let z become x+h 
os Sfi=a(e-8)2 
Expanding by the binomial 


Ee? A Caga! asi) 


aon AE my pli—l n—-2 },2 n—3],3 
=2*% +na"At+ hei eh? + ae ze shs + & eo, 
dy 
° Siete n—l 
; qe 


From this it is manifest that 


The first differential co-efficient of any power of a simple 
Algebraic quantity is found by multiplying the quantity by the 
index of the power, and then diminishing the exponent by unity. 


Ez. 1. y= x" ay =a" {a 
2. ya arte a = @ (p+q) zPti! 
3. y = at+x-4 ie == — gat) 
4. y= Gis _ = eG) 


INTEGRAL CALCULUS. 


The object of the Integral Calculus is to discover the pri- 
mitive function from which a given defferential co-efficient has 
been derived. : 

This primitive function is called the integral ofthe proposed 
differential co-efficient, and is obtained by the application of 
the different principles established in finding differential co- 
efficients and by various transformations. 


& 12 


_ 


170 DIFFERENTIAL AND 


When we wish to indicate that we are to take the integral 

of a function, we prefix the symbol §, Thus if 
y = ax’ 

We know that dy = 4aa‘dx 

If then, the quantity 4az* dx be given in the course of any 
calculation, and we are desirous to indicate that the primitive 
function from which it has been derived is ax‘, we express this 
by writing 

8 4az* dx = azx* 


The characteristic §$ signifiesintegral or sum. The word 


sum, was employed by those who first used the differential and 
and integral calculus, and who regarded the integral of 2™dx 
as the sum of all the products which arise by multiplying the 
mth power of z, for all values of z, by the constant dz. 

When constant quantities are combined with variable quan- 
tities by the signs + or — they disappear in taking the differ- 
ential co-efficients, and therefore they must be restored in 
taking the integral. 

Thus, if y = az +b 


or, y = ax* —b 
or, y = ax’ 
In each three cases equally 
dy = sax’ dx 


Hence in taking the integral of any function itis proper al- 
ways to add a constant quantity, which is usually represented 
by the symbol C. Thus, if it be required to find the integral 
of a quantity such as 


dy = 3ax’* dx 
yY = § 8ax' dz 
=azx'+C 


where C may be either positive, negative, or 0. We cannot 
determine the value of C in an abstract example, but when 
particular problems are submitted to our investigation, they 
usually contain conditions by which the value of C can be as- 
certained. 

By reversing the principles established for finding the dif- 
ferential co-efficients, or differentials of functions, we shall ob- 
tain an equal number of rules for ascending to the integrals 
from the derived functions. Recurring therefore to these we 
shall perceive that 


1. The integral of the sum of any number of functions is 
equal to the sum of the integrals of the individual terms, each 
term retaining the sign of its co-efficient. Thus, if 


INTEGRAL CALCULUS. © 171 


dy = 4ax’* dx + 8bx? dx — 2bx dx + dx 
y =§ 4az* dx + $ 3bx* dr—$ 2 bz dx + §dx+C 
II. Since, if | 


y = a™ 

dy = maz™ az 
it is manifest that 

The integral of a function raised to any power is obtained 
by adding unity to the exponent of the function, and by divid- 
ing the function by the exponent so increased, and by the differ- 

ential of the funciton. 
ee 1. dy = ax" dx 
_ axl 
~ n+l 


Ex. 2: dy = de 


= ux dz ' 
ax er 
Paps 29) 
a 
Re Pinay che 
Ez. 3. dy = (a+ 2x)" dz 
_ (a + x)etl 
n+l 
dy = seldaie. dx 
I= (a+a) 
se (Grit) oar 
1 


+C 


+6 


7 =~ ~(@—l) (a+2z)-2 
This rule applies to all functions of the form 
dy = (a + ba™)™cx"7! dx 
for these can all be reduced to the form az™ dz. 


Scholium. 

The coincidence of the results obtained by the notation, 
used in Chap. Hl, with those determined by the Calculus, in- 
dicates that the same principles serve as the foundations of 
both ; which, as was observed at the close of Chapter I, are 
the following, viz., that when any series of quantities vary in 
arithmetical progression, the sum of the series as well as their 
powers, or products, with a constant quantity, or with another 
arithmetical series, may be measured by the sum of the pro- 
ducts of the maximum values + the minimum values + four 
times the products of their medium values multiplied by } of 
the magnitude, representing the number of the series. 


172 DIFFERENTIAL AND &c. 


Any series of numbers which are not subject to these con- 
ditions, cannot be made the subjects of accurate numerical 
determination or calculation, by the Calculus ; hence, we 
may always determine, by construction, whether any geome- 
trical magnitude, or algebric quantity, is susceptible of accu- 
rate development numerically in terms of given quantities, 
One advantage possessed by the investigation of geometrical 
subjects, by the principles contained in Chapter If, of this 
subject, is, that the notation there used, is more elementary 
than that of the Calculus, and expresses the conditions of 
geometrical subjects in a more obvious and inteligent manner, 
and is more direct in its results. 

Hence, instead of pursuing the method of defferentation, 
we express the conditions of quantities depending on variable 
factors according to their several conditions, or organization, 
and proceed to integrate or sum up the series of functions, 
and determine the magnitude of the production, according 


to the principles therein contained. Hence S§ x’ expresses 


definitely the function of the series of squares of a series of 
variables, and if this series be drawn into a constant quantity 
x, and integrated, we shalt have according to our notation 
iz’. 

" If this is made equal to uw, or if we have u = 12°, its differ- 
ential is 2 =z", which, being again integrated by the cal- 
culus, we have again 32°, showing similar results, from differ- 
ent considerations. The notation by the calculus is arbitrary, 
but by this inductive ; and hence, by its intimate connection 
with the principles of the calculus, may serve to render that 
subject more obvious. 

There are cases, however, for which the calculus is more 
particularly adapted than the notation here referred to; such 
as drawing tangentsto, and rectifying curve-lines. This is 
accomplished by that science in a manner the most elegant 
and complete ; but in relation to surfaces and solids, nothing 
can be more complete or satisfactory, than the discussions 
which we have introduced. 


OF THE VIRTUAL CENTRE. 173 


CHAPTER IV. 


On the centres of surfaces and solids, the method of finding 


them, &c. 


Article 1. The centre of surfaces and solids is susceptible 
f two considerations, that of magnitude and distance. Hence, 
we have the centre of aggregation and the virtual centre. 

2. The centre of aggregation or centre of magnitude of any 
plane surface, is that point through which, if a line is drawn 
in any direction in the plane, it shall divide the surface 
equally. 

3. The centre of magnitude of any solid is that point, 
through which if a plane is passed in any direction, the plane 
shall divide the solid equally. 

4. The virtual centre, or centre of gravity of any plane 
is that point through which, if a line be drawn in any direc- 
tion in the plane, the sum of all the points, on one side, drawn 
into their distances from the line, shall be equal to the sum of 
all the points on the other side drawn into their respective dis- 
tances from the line. 

5. The virtual centre of any solid is that point through 
which if a plane be passed in any direction, the sum of the 
points, on one side, drawn into their distances from the plane, 
shall be equal to the sum of al! the points on the other side, 
drawn into their distance from the plane. 

6. The virtual centre between any two points is evidently 
in the right line connecting the two points, and equidistant 
from each. 

-Hence the virtual centre of a right line, is in the middle of 
that line, oris in the centre of magnitude of the line. 

7. In a system of points, their virtual centre has respect to 
the magnitude of the lines drawn from the several points to 
such centre. 

But the centre of magnitude, has regard only to the number 
of points. 

8. If there be drawn froma system of points, lines perpen- 
dicular to a given base, the sum of those lines divided by 
their number, will give their average length ; whichis = to 
the distanee of the virtual centreof the system of points from 
the base. 


174 _ METHOD OF FINDING 


Let AB be a given base line, and 
let a, b, c, d, e, &c., be a system of 
points ; if perpendiculars al, b2, c3, 
&c., be drawn from the points to 
meet the base line AB, and if the 
sum of these lines be divided by their 
number, the quotient will give their 
average length, or the distance AC, 
from which, if a line CD be drawn, 
parallel to AB, it shall pass through the virtual centre of the 
system of points. 

For if the line CD is drawn, according to the conditions ex- 
pressed in the proposition, the portions of the lines cut off 
beyond the line CD, are sufficient to extend those which fall 
short of that line, to the line CD. Hence, the line CD passes 
through the virtual centre of the system, for the lines drawn 
from the points on both sides of the line are equal. 

Hence, if two base lines 8 
AB, GH are drawn, not pa- 
rallel to each other, and if 
the system of points are 
connected to both of the 
bases, and we proceed, as 
in the proposition to draw 
CD parallel to AB, and IL 
parallell to GH, the inter- 
section F’, of the lines CD, 
and IL will be the virtual centre of the system. 

If the base line AB, should pass through the system of 
points, so as to leave one portion of them on one side, and 
another portion on the other, then the lines connecting the 
points to the base must be estimated on one side positive, and 
on the other negative ; and the virtual centre will fall on one 
side or the other, as the positive or negative signs predomi- 
nate. 

9. If, instead ofa system of points, the virtual centre of a 
system of parallel lines is required. Let a base line be drawn 
parallel to those lines, and because the lines are to each other, 
as the number of equidistant points, arbitrarily taken in each ; 
hence, if we take the product of the several lines into their re- 
spective distance from the base, the sum of these products 
divided by the sum of the lengths of the lines, will deter- 
mine the distance from the base, through which, if a line is 
drawn parallel to the base, it shall pass through the virtuai 
centre. 


KH WNWRraAn tag? 


THE VIRTUAL CENTRE. 175 


10. The virtual centre of a surface may be found by sup- 
posing parallel ordinates drawn across its surface, and assuming 
those ordinates, as representing the surface, and computing 
the distance of the virtual centre of the ordinates from a 
given base ; and if from the properties of the figure, it cannot 
be readily discovered what part, of the line passing through 
the centre, is occupied by the centre, then another base may 
be assumed, making an angle with the former, and if ordinates 
are supposed to be drawn across the figure, parallel to this 
base, another line may be found parallel to this base, passing 
through the centre ; hence, the intersection of these two lines 
will be the centre required. 

11. If AQ bea base line drawn through B__ Lv dt 
any point, as suppose the vertex of any ||x ==” 
body, or figure QBD, and if @ denote any 
ordinate EF of the figure, d = AG, its dis- 
tance from the base line AQ, and S = thea 
sum of all the ordinates, or the whole figure 
QBD ; then the distance IC of the virtual centre from AQ, 1s 
denoted by the (sum of all the ad) + S. 


PROBLEM I. 
To find the centre of a triangle. 


Through the vertiee C of the triangle 
draw a base line CD, parallel to AB, let 
an indefinite number of ordinates ab, or ’ 
z, parallel to DC, be conceived to be drawn 
across the triangle each of which we 
may conceive to be drawn into their re- 
spective distances be, &c., from the line 
CD, which may be represented by a se- 
ries of x; hence, if the series of z, whose F E 
magnitude is x, bedrawn intothe series of zx, the product may 
be represented by a pyramid, whose base ABEF is = zz 
and whose altitude is AC, or x; hence, } AB X BD X BD= 1 
AB xX BD? or xzx,=417z,=the sum of the products, but (Art. 
2) the sum of the products + S, the sum of the series of or- 
dinates, or series of z, whose measure is zz, is equal to the 
distance of the centre, from the line CD. The sum of the 
series of ordinates may be expressed by half the rectangle of 
AB x BD, or}ze. Hence, by (A sista 

x BD, orize. Hence, by (Art. 11.) AB xX HD 


&. 


= 2AC ="2DB or 4232 + $zr = 2x, that is the virtual cen- 
tre Is some where in a line ab 2 of the distance from the ver- 


176 METHOD OF FINDING 


tice C of the triangle to the base AB, in like manner, we shall 
find it also in a line GH, 1 of the Retanch from the vertice B 
toward the side CA ; heater it must be in the intersection of 
the two lines. 


Problem 2. Let it be required to find the virtual centre 
of a trapezium, ABCD. 


First, let DC, parallel to AB, be taken as Dees FG 
the base and imagine an indefinite number 
of equidistant ordinates ab, drawn across the 
figure, parallel to the base, and if these are 


severally drawn into their respective dis- 


tances from the base DC, we shall have con- 


structed a wedge, whose base is equal to AB, AEE E==—SB 
drawn into EF, and whose altitude is equal ne 
to EF, 


Let z = AB, and z’ = DC, then will the series of ordinates 
be represented by z..z’ let EF = z, then will the series of de- 
creasing distances be represented by z. 

Then, we have the solid generated by the production of 
E X(Z..2)) asd Be eles iar Lae x(2z +z’), let this be divided by 4 
aheras) 
3(z+2') 
tual centre I from the line DC. 

Now, in order to find in what part of the line Ab is the cen- 
tre, we may proceed as before to construct a quantity on CB, 
DG, or any line parallel to CB, consisting of ordinates EF, 
&c., across the figure, drawn into their respective distances 
from the bases, and dividing the quantity thus constructed by 
the sum of all the ordinates, or the area of the trapeziums. 

Or, we may divide the trapezium into the triangle ADG, and 
the parallelogram GDCB, and proceed to find the virtual. 
centres of each of these figures ; and it is then evident, that if 
the distance between the two points, thus found, is divided in. 
the alternate ratio of the two figures, the result will determine 
the centre required. 

Thas, the distance he of the triangle, from the side DG is } 
aH, (Prob. 1,) and the distance ne of the centre of the pa- 
rallelogram from the line DGis evidently = } nb. 

Let ¢ = the area of the triangle, and p = that of the paral 
lelogram, then as ce the distance of the two centres, is to t + p, 
so is p, to the distance el, of the vertical centre of the trapezium 
from the point e. 


(z + 2z')z, and we have = the distance of the vir- 


THE VIRTUAL CENTRE. 177 


Any rectilinear figure may, if necessary be divided into 
triangles, and the virtual centres determined for each, and 
then finding the common centre of every two of these, till 
they are all reduced to one only, which will be the virtual 
centre of the whole. 

Problem 3. To find the virtual centre of a segment of a 
circle ABD. 

Let parallel ordinates ab, &c., be drawn 
across the segment parallel to its base AB, 
and let each of these ordinates be con- 
ceived to be drawn into their respective ¢ 


distances from AB; then by their proper- A E B 
ties we shall have generated an ungula ABCF; divide the soli- 
dity of this ungula by the area of the segment, and the quotient 


is the distance EF of the virtual centre from the chord AB. 
D H 


Or we may let GH be the origin, G 
passing through the vertice D of the 
segment, parallel to AB, and if the 
ordinates ab, &c., are drawn into 
their distances from this line GH, or 
the vertice D, the product will be the 
ungula ABDL, which is the compli- 
ment of the former ungula, and the distance DI of the centre 
will be found by dividing the solidity of the ungula by the 
area ABD. 

If we make the origin at the centre 
of the circle, we shall have the ungula 
ADBRCF, which is a segment of the un- 
gula GDHF, and the solidity of this seg- 
ment, divided by the surface of the seg- G O 
ment ABD, will give the distance OI, of the centre required. 


Problem 4, Let it be required to find the virtual centre of 
a sector of a circle CADB. 

Let C be the origin, and if an inde- 
finite number of equidistant ordinates 
db, parallel to the chord AB, are drawn 
into their respective distances Ce, &c., 
we shall have produced the sectoral un- 
gula, FADBGC, which if we divide by 
the surface, CADB, we shall have the 
distance Cl of the centre. 

Let r = CD the radius of the circle, and the chord AB = c 
then will rc = the convex surface AFDGB of the ungula, and 
ir’c = the solidity of the ungula. Let 7’ = the arc ADB, 


178 METHOD OF FINDING 


then we have the area CADB = i1’r; hence the distance CI 
prin bg 2rc 
= la'y — 37? 

Resolving this into a proportion, we have, 37 : 2r::c: the 
distance CI. Hence, generally, the distance of the virtual 
centre of any sector of a circle is = the fourth proportional to 
three times the arc of the sector, twice the radius, and the 
chord of the arc. 

Let CE = V(r? —1c’,) hence we have !7°c — 1c* = the so- 
lidity of the pyramid ABGFC ; subtract this from the sectoral 
ungula, and we have j, c* for the solidity of the segment 
ABGFD, divide this by a, the area of the segment ABD and 
we have the distance of the virtual centre of the segment 


3 


an i 
T2© 


from the centre C of the circle = 


Problem. 5. To find the virtual centre of the parabola. 
Let its distance be estimated from the vertex. 

. Phe equation to the parabola is y?= ./pa, if this be drawn 
into a series denoted by z, the surface of the parabola will be- 
come y x/dz. 

Let this be drawn into a series Z, and we have zz/pz, and 
since the quantity p will not affect the result, it may be omitted, 


. . : 1 3 
and we have by removing the radical sign x12 = 2.22; let 


this be divided by zzz = 2 23 and we have 


ZU 
ertex. 

Problem 6. To find the distance of the virtual centre of a 
semi-parabola from the axis. 

Its equation considering the origin at the extremity of the 
axis 1s y= (d—z)’+22(d—zx);d=the maximum value of =the 
axis ; or its equation may be expressed 2? + 2zz. This 
being put into a series, whose measure is a, represented 
by the base, we shall have its area, which may be ex- 
pressed ax’+2ax % = dy, which being integrated, we have 
4 ax" + § az’ = 2az’. 

Let the expression a2” + 2axx, be drawn into z, z being 
= the base, and we have aza* + 2azzz, let this be inte- 
grated, and we have 4(;zx7 + 12x) 1a = la(11z2*) = [7 
Gea. 


2o2 , 
- j = 3x” = the distance of the virtual centre from the 
3 


7 == 2a= the distance of the virtual centre from 


the axis. 


THE VIRTUAL CENTRE. 179 


Prob. 7. Let it be required to find the virtual centre of a solid. 
If the solid be a pyramid, its solidity may be expressed x?z. 
If we estimate the centre in relation to its distance from the 
vertex, we may draw this quantity into another series of z, 
and we have xz, let this be divided by z?z, which represents 
the base of the latter construction, and we have 
ex 4x é 
—~— = “=z, that is the vertical centre is 2 the distance 
De LE 
from the vertex of the pyramid to the base. 


Problem. 8. To find the virtual centre of a vertical seg- 
ment, of aspherical revoloid, or the virtual centre of a seg- 
ment of a sphere. 

The equation of the reveloid is 4y2 = 4dz—z?, let this be 
drawn into a series of z,and we have 4rxz (d—z), the segment, 
x being the altitude of the segment, and d the diameter of the 
sphere, the value of this is 

1(2dx2—2zxr3) + 1(4dx?—2zx*) = dr—?2z?. 
Let the former series be drawn into another series of z, and 
its value will be 4%(dx2+Zx? = 2dx3—1 24; 
2dxz3—1r4 = 4dr—3x? 
hence, ——-——- = ——— —— 
di 47% 6d—4x 
centre of the segment from the vertex. 

This construction will serve for a segment of a sphere, a 

spheroid, a spherical, or an elliptical revoloid. 


=the distance of the virtual 


Problem. 9. To find the virtual centre of a vertical parabo- 
lic revoloid pyramoid, or of a parabolic conoid. 

The equation to the revoloid is y’=pz, let this be put into a 
Series, whose measure is the axis, and we have its solidity 
apx; let this, as a base, be also drawn into z, and we 
have pxz” equivalent to }pz° ; 
xpx is equivalent to } p2’ ; 

1 3 
Me 


1 
g Pe 


hence ; = 2x = 2% the distance from the vertex to. the 


base. 


Problem. 10. To find the virtual centre of an hyperbolic 
conoid, pyramoid or vertical revoloid. 
c2 ; c? 
Its equation is 4y? = 4 7 (dx + x*) : since 4a 


stant quantity, and a factor of the whole expression, it may be 
omitted, without affecting the result ; then the expression will 
become dx + x?. Let this be drawn into a series of the form 
z(d+xz)_, which expresses the segment, whose value is, 
3(dx+a?2) +} (idx+}2*)=fa(8de+2z?) = 5 (3dz*+2z') 


is a cone 


180 METHOD OF FINDING 


Let the former series be drawn into another series of x, and 
there will be produced x(d + x)x? == 12(dx? + x3) + tex 
(jdz? +123)4=12(2dz*+112°)=1(2dz*+ 112°), hence, we have 

I 3 Lint 2 
4Qde% + Liz), Ade. Bm 

Riddatp enw tenes the distance of the centre from 
the vertex. 


Problem. 11. To find the virtual centre of the arc of a 
circle. 

The virtual centre of an are of a circle, is the same 
in reference to the centre of the circle, as that of the segment 
of arevoloidal curve, whose conjugate diameter is the same 
as that of the circle, and the base of whose segment is equal 
to the given arc of the circle. 

For, if a_revoloidal 
curve AEB circumscribe 


atu | 
the semi-circle DEH, i Use 
and if any chord FG, be /| \ 
produced to fg, soas to | 


meet the curve, the ordi- 
nate fg will be equal to the arc FEG. Draw across the seg- 
ment fEgf, equidistant ordinates, perpendicular to fg, and 
they will represent the distance of the several points in the 
arc, from the line fg, since they are supposed to be drawn from 
points equidistant from each other on the line fg, or the arc 
FEG; and, since if there is an infinite number of ordinates, 
they may be regarded as the area of the segment, it there- 
fore, follows that if this area is divided by the arc, the quotient 
is the distance of the centre, from the line FG ; also, if the 
area f EgHD is divided by the arc FEG or line fg, the quo- 
tient is the virtual centre of the arc f Kg from the axis DH. 

Let c = the chord FGa = the arc FEG, and r = CE, then 
will cr = area Df EgH, (Prop. Ill, Cor. 4, B. III.,) hence, we 


ler 
have aa the distance EI of the centre. 


This is also the virtual centre of the segment f Egf of the 
revoloidal curve, and it may also be shown that if a series of 
equidistant ordinates to the axis Cl, are drawn through the 
revoidal surface, or any segment of it parallel to fg, the series 
of ordinates so constructed, drawn into their several distances 
from any given line, parallel to such ordinates, will determine 
the virtual centre of the segment fgef, or of the arc FEG, by 
proceeding as before. 

Art. 12. If it be required to find the virtual centre of the arc 
of an ellipse, a parabola, or an hyperbola, it may be done in 
4 similar manner, by taking a portion of the surface of the 


THE VIRTUAL CENTRE. 181 


elliptic, parabolic, or hyperbolic reyoloid, and proceeding as 
for the arc of the circle, which also gives the virtual centre 
of the segment of the revoloidal surface pertaining thereto. 

Problem 3. To find the virtual centre of the surface of a 
solid. 

Let the proposed surface be the convex surface of a pyra- 
mid ABCDYV ; and because any portion of the convex sur- 
face, included between any two sections, by planes parallel to 
the base, is proportional to the portion of a vertical triangle 
through the pyramid included between the same planes; it 
follows that the vertual centre of the convex surface, is the 
same as that of the vertual triangle ; hence the same process 
will determine both. If it is required to find the virtual cen- 
tre of the whole surface of a pyramid, including its base, we 
have only to imagine an infinite number of ordinates to be 
drawn across the several triangular sides parallel to their se- 
veral bases, and also a similar series of parallel ordinates 
across the base, and if each of these ordinates are severally 
drawn into their respective perpendicular distances from the 
vertex of the given pyramid, we shall have produced, as many 
new pyramids AEFCQ, whose bases R 
ACFE, ABHG, &c, are, severally 
equal to the bases of the sides of the 
pyramid multiplied by IA, the dis- 
tance of the base from the vertex, as 
the given pyramid has sides, and also 
a prisn ABDCQIKR, formed by 
drawing every line in the base, or the 
whole surface of the base into the 
distance of the base from the vertex. 
And the sum of the imaginary solids so generated, divided by 
the whole surface of the pyramid, will give the distance of 
the virtual centre from the vertex. 

Let AB=z and if the pyramid is generated from 2’ or hz’, 
or if the base of the pyramid is a square, the perimeter of the 
base will be 4x ; and since each pyramid ACF EQ, is equal to 
3 the prism ABCDQRKI: hence the four pyramids generated 
by the series drawn into the four sides of the base are = 4 the 
prism, and if h = the altitude IA of the pyramid, ha’?+4ha?= 
thx" = the sum of the four pyramids + the prism ABDCQIKR ; 
the surface of the given pyramid is = 2’+92/(h?+32’) 

Hence we have, aha Thx 

4+ Qn /(h?+4a%7 242 Vh?+42? 
equal the distance of the virtual centre from the vertex. 


182 ON THE RELATIONS OF MAGNITUDES. 
BOOK. VII. 
CHAPTER V. 


ON THE RELATIONS OF LINES, SURFACES, AND SOLIDS, 
GENERATED BY MOTION. 


* Tue capacity of any solid generated by the motion of a 
surface perpendicular to itself, is measured by the generating 
surface drawn into the distance moved ; which distance is al- 
ways equal to the distance passed through by the virtual cen- 
tre of such surface. 

If the motion of the generating surface is such, as that it 
always maintains a parallel position, and moves in a direction 
perpendicular to itself, the proposition is sufficiently manifest. 


BS) \G 


Let now the rectangle ACBD revolve about R 
the side BC, which remains fixed, and the 
product will be the cylinder DF’, whose solidity 
is equal to the surface ACBD drawn into the 
circumference PK, described by the virtual 
centre K, of the plane, which centre is in 
this case also the centre of aggregation. 


If a right-angled triangle ABC revolve about the perpendi- 
cular BC, so as to describe the cone ABD, P B N 
this is also measured by the triangle ABC 
drawn into the circumference FL, described 
by the virtual centre of the triangle. The 
virtual centre of the triangle we have 
shown to be situated at the point F, on the 
line BE, from the vertex bisecting the base 
at a distance from B =? its length. Let 4’ 
the surface ABC be multiplied by the cir- 
cumference described by the centre F’; and since the radius 
FG=2EC, hence the circumference h}G=2 the circumference 
EC, and because the triangle ABC=1Lits circumscribing rectan- 
gle ADBC, which generates a cylinder ADNM, the generating 
surface of the triangle ABC drawn into the circumference, is 
equal to one-third the cylinder generated by the rectangle, or 
one-third the rectangle drawn into the circumference EC, as 
it ought to be. 

And in general, let any plane figure be revolved about any 
line or axis without the figure, but always in the same plane, 


GENERATED BY MOTION. 183 


and the solid generated will be measured by the generating 
surface drawn into the are described by the virtual centre of 
the surface. 

Let AFHD be a solid generated by 
the plane ABD; through C, the virtual 
centre of which, draw DCAE, perpen- 
dicular to the axis of rotation, and meet- 
ing HGFE in E, let an indefinite num- 
ber of parallel ordinates, ef, 2k, &c., be 
drawn across the generating surface, 
parallel to the axis about which it re- 
volves; and the solid generated is equal 
to all of those ordinates, drawn into the 
distances passed through by each; viz., 
the ordinate drawn across the point 
Axthe arc AF+the ordinate ab drawn 
through cXthe arc CG+, &c., through © 
the whole series. And because EA, 
EC, ED, &c., are as the arcs AF, CG, DH, &c. Hence 
EhxXef, and Elxik, &c., are as phXef, and rlxik, and be- 
cause EC drawn into all the ef, ik, &c., is equal to all, the 
EhxXef, Elx tk, &c., it follows that CGXall the ef, ik, &c., is 
equal to all the phxXef, rlxik, &c.; or that the solid ABDHF 
is equal to the generating surface ABDe drawn into the line 
described by the virtual centre of the surface. 


Cor. 1. Hence, if any curve or any line be made to revolve 
about any axis exterior to such curve, but in the same plane, 
the surface described by its motion will be equal to the line or 
curve drawn into the distance passed through by the virtual 
centre of such line or curve. 

For, let the perimeter of the figure generating the solid 
above, be the generating line, and let us suppose its virtual 
centre the same as before; let every point in this perimeter 
be reduced to the line AD by means of perpendiculars thereto ; 
and the figure generated by its revolution about the axis, is 
equal to all the ph, rl, &c., described by every point; but 
we have seen that all the pA, rl, &c., are as all the Eh, El, 
&c. ; and since the sum of all the EA, EJ, &c., is equal to as 
many times EC, therefore the sum of all the ph, rl, &c., is 
equal to as many times CG, or equal to ABDeXCG, that is, the 
surface described by the perimeter ABDe, is equal to ABDe 
drawn into the line described by its virtual centre C. 


Cor. 2. From E draw EIKL, cutting the upright prismatic 
figure erected on the given base ABD, so as that any perpen- 


184 ON THE PRODUCTION OF MAGNITUDES. 


dicular AI may be equal to the corresponding arc AF. Then 
will the figure AILD be equal to the figure AFHD. 

For, by similar figures, all the AF, CG, DH, &c., are as all 
the AI, CK, DL, &c., each to each; and as one of each are 
equal, therefore they are all equal, each to each; viz., all the 
Al, CK, DL, &c., equal to all the AF, CG, DH, &c.; that is, 
the figure AILD equal to the figure AFHD. 


Cor. 3. Through K draw MKNO; then the figure ANMD 
will be equal to the figure AIKLD, or equal to the figure 
AFHD. 

For, by the last corallary, AFMD is equal to the figure de- 
scribed by the base AD, revolving about O, till the arc de- 
scribed by C be equal to CK; which, by the proposition, is 
equal to ADXCK, or ADXCG. 


Cor. 4. Hence, all the upright figures AQKRD, AIKL, 
ANKMD, AKPD, &c., of the same base, and bounded at the 
top by lines or planes cutting the upright sides, and passing 
through the extremity K, of the line CK, erected on the vir- 
tual centre of the base, are equal to one another; and the va- 
lue of each will be equal to the base drawn into the line CK. 

Hence, also, all figures described by the rotation of the same 
line or plane about different centres or axes, will be equal to 
one another, when the arcs described by the virtual centre 
are equal. But if those arcs be not equal, the figures gene- 
rated will be as the arcs. And in general, the figures gene- 
rated, will be to one another, as the revolving lines or planes 
drawn into the arcs described by their respective virtual 
centres, 


Cor. 5. Moreover, the opposite parts NIK, MLK, of any 
two of these figures, are equal to each other. 


Cor. 6. The figure ASPD is to the figure APD, as AS to 
CK ; for, by similar triangles, they will be as AD to AC. 

For ASPD is equal to ADX AS, and APD equal to AD 
xCK. 


Cor. 7. If the line or plane be supposed to be at an infinite 
distance from the centre about which it revolves, the figure 
generated will be an upright surface or prism, the altitude be- 
ing the line described by the virtual centre; so that the base 
drawn into the said line will be equal to the base drawn into 
the altitude, as it ought for all upright figures, whose sections 
parallel to the base are all equal to each other. 


BY MOTION. 185 


Scholium. If a right line, or parallelogram, revolve about a 
line perpendicular to the length, there will be described a ring 
either superficial or solid; and as the virtual centre of the 
describing line, or parallelogram, is also the centre of magni- 
tudes, it follows, therefore, that such surfaces or solids, are 
equal to the generating magnitude drawn into the distance 
passed through by the centre of magnitude. 

When the centre of rotation is in the end of the line, the 
line will describe a circle whose radius is the said describing 
line, and whose circumference is double the circumference 
described by the virtual centre; consequently, the radius 
drawn into half the circumference, will be the area of the 
circle. 

If a semi-circle revolve about a diameter, and describe the 
surface of a sphere, then will the surface of the sphere be 
equal to the revoloid arc X by the circumference described by the 


: cr ’ ‘ 
virtual centre = ee See Xa = 2rev = the circumference into 
z 
the diameter. 
And for the solidity of the sphere, we shall have the dis- 
3 


tance of the virtual centre equal , where d is the diameter, 


12a 
and a the area of the segment ; let twice this distance be mul- 
2d°r a’ h 

ign tg the so- 
lidity of the sphere, as before found in the Elements of Ge- 
ometry. 


tiplied by +, and also by a; and we have 


For the solidity of the parabolic spindle: putting b = the 
base, and a@ = the altitude, or axis of the generating parabola. 

We have found that 2a is the distance of the centre of gra- 
vity from the base, and consequently *%a7 = the line describ- 
ed by the centre of gravity ; but 2ab is = the revolving area ; 
therefore 1Sar x 2ab + the 22a’br will be the content, which 
is ;% of the circumscribed cylinder. 


For the paraboloid. Making the notation as in tne last ex- 
ample, and making 2 = the area of a circle, whose diameter 
is 1; 3b will be the distance of the centre of gravity of the 
semi-parabola from the axis, consequently 3b X 8n X 2ab= 
2ab’n = the solidity = half the circumscribed cylinder. 


13 


MENSURATION. 


Having, in the elementary parts of the work, introduced 
such subjects of mensuration as depend on principles therein 
discussed, it only remains for us now to present the higher 
branches of the subject, or such subjects in mensuration, as de- 
pend on the higher branches of geometry. 

The subject of mensuration admits of three general divi- 
sions : lines, superficies, and solids ; but since the mensuration 
of lines is so intimately connected with that of surfaces, we 
shall make but two general divisions, termed superficies, and 
solids. 


PART I. 
MENSURATION OF SUPERFICIES. 


PROBLEM I. 

To find the area of a segment of a circle. 
CASE I. 

When the arc, sine, and radius are gwen. 


Rutz.—Multiply the difference between the are of the 
segment and its sine by half the radius. (Prop. XII, B. IV.) 

Let «’ = the arc, s = the sine, 7 = the radius, and A the 
area of the segment; and A = 1a’r—tsr. 


Ex. 1. What is the area of a seg- 
ment AE, whose arc AE is 2,09438, 
and whose sine ES = 1,73205, the ra- 
diliss— 22,4 

2.094388 


the area of the segment AE. 


D 
Eix. 2. What is the area of the segment EADB, whose arc 
EADB = 4.18878 and sine ES = .86602 ? 
In this example because the sine is considered negative, by 
Trigonometry, the arc being greater than that of a semi-cir- 
cle, we shall have by the rule 


4.18878 ) . ; 
+ .86602 | 2” = 1.66138 = the area required. 


MENSURATION, ETC. 187 : 


Ex. 3. What is the area of the segment whose arc is 
6.9813, and whose sine is 6,4278, the radius being 10? 


Scholium. If the are and radius, or the sine and radius 
only are given, the other parts may be taken from the table of 
natural sines, and the area of the segment calculated by the 
rule. The arc of any segment less than a semi-circle may be 
found approximately by formula 3, (Prop. IX, B. IV.) 4¢ = 
J/(4vur-+1s’) —4s. Where v is the versed sine or height of 
the segment, r the radius, and s the sine of the half arc, or 
the 1 chord of the segment, x being = the arc of the segment. 

It will appear that this gives the value of the arc to a 
great degree of exactness when the segment is small. 

Let us see how near the truth this comes for a semi-circle. 

In this case, the sine and versed sine are each equal to the 
radius, which suppose = 1. 

Whence we have iz = V41—4 = 1,56155 

And x = 3.12310 
the true number being 3.14159 

The difference of which is .01849, the error in a segment 
= the semi-circle. 

Let us assume that the error in any smaller segment, is pro- 
portional to the sixth power of v or v’, then if we correct this 
by deducting v’ x.01849 therefrom, the result will, in this case 
be correct, and if our hypothesis is correct, it will give the 
proper result for any smaller arc. 


CASE II. 
When the arc, cherd, versed sine, and radius are given. 


Rute.—Multiply the difference between the chord and arc 
by half the radius, to which add half the product of the chord 
and versed sine. 

Investigation. Let ABD be a segment; this is <7 
composed of the segments AD + DB + triangle »f/ \ | 
ADB, but the segment AD = DB = (are AD — AF) linn 
Xir; segment AD+DB = (are ADB —chord AB) & 

X2zr and triangle ADB = ;(ABx DF). 

Ex. What is the area of the segment ABD, whose arc 
ADB, is 10,4719, and whose chord AB=10, the versed sine 
DF or height of the segment =1.3398 ? 

10,4719 10 X 1.3398 


segment. 


=,9058 = the area of the 


> 


188 MENSURATION 


CASE Hi. 


When the radius of the circle, and the degrees of the arc only 
are given . 


Rure.—Find by Frigonometry, or by the table of natural 
sines, the sine of the given are for a circle whose radius is 1, 
observing that the same sine answers for an are and comple- 
ment, multiply this by the given radius, which gives the sine of 
the given arc. Then say, as 180° is to the given arc, so is + 
to rt’, so is the semi-cireumferenee of a circle whose diameter 
is 2, to the length of the given arc; then proceed as in Rule 1. 


Ex. I. What is the area of the segment AK, (see diagram 
to Case 1,) whose arc AK: = 60°; the radius EC being 1? 
/ (EC? — SC’?)=ES= ¥(r——-ir)= v (4) =.86602=s 
And 180° : 60°: : 7:4’: : 9.14159 : 1.04719 = the arc AE 
=, Hence, (1.04719 — .86602) x3=.09058 = the area of 
the segment AE. 


Ex. 2. What is the area of the segment EBF, whose arc 
EFB is 120°, the other quantities remaining the same as be- 
fore? 

The sine of the arc AE is also the sine of the arc EFB. 
The are EFB=2 arc AE=2.09438; hence (2.09438 — 
86602) X+=.61418 = the area of the segment EBF. 


Ex. 3. What is the area of the segment EADB, whose arc 
EADB is 240°, the other quantities remaining the same ? 

As 180° : 240: : 3.14159: 4.18878=the arc EADB. 

Since the segment EADB is greater than a semi-circle, its 
sine, ES, is considered negative by Trigonometry, we have 
(4.18878 +-,86602) X $=2.52745 = the area of the segment 
E ADB. 


Ex. 4. What is the area of the segment ADBFE, whose 
arc is 300° 7 A 


Scholium. The difference of the segments EBF, and the 
segment AF is equal to the sector ACE ; the segment ADC ~ 
segment EBI — sector ACE + triangle ECB. 


OF SUPERFICIES. 189 
CASE IV. 


When the chord of the segment, its height or versed sine and 
radius are given. 


Rutrsr.—As the radius is to half the | 
‘chord, so is twice the difference of the 
versed sine and radius, to the sine of / 
the arc of the segment; divide this by Fr 
the radius, reducing it to the sine of an 
arc of a circle, whose radius is 1. 

Then in the table of natural -sines, 
take out the arc answering to that sine 
in degrees, and proceed as in Rule II], 


to find its length ; then proceed as in rule Ist., to find the area ~~ 


of the segment. 

Investigation. Inthe right angled triangle FAB, we have 
FA—2CH, and because the triangle ASB is similar to FAB 
or CEB—hence, CB: EB:: FA: AS. 

Or without finding the sine AS, of the arc ADB, proceed te 
take out from a table of natural sines the arcs AB, DB, an- 
swering to AK, BE, the sines of those arcs respectively ; and af- 
ter finding their lengths as in Rule ILI, proceed by Rule II, to 
find the area. 

Ex. 1. What is the area of a:ssegment ABD, whose chord 
AB=17,3205, and whose height ED=5, the radius being 107 

In this example we have FB=20, AB=17,3205 and AF== 
(10 — 5)2—10, to find AS. 

Or, we may make the triangle CEB, whose side CB—10, 
EB—8.6602, and CE = 5, and FA=2CE to find AS, by the 
rule. 

Hence, 10 : 8.6602 : : (10 — 5)2 : 8.6602 — the sine; and 
8.6602— 10—.86602 = the tabular sine: the arc answering 
thereto is that of 60°, but this segment being greater than a 
quadrant, the arc must be the supplement of 60°120°. Then 
180° : 120° : : 3.14159 : 2.09439 = the length of the arc of 
120° in a circle whose radius is 1. 

Hence, 2.09439 x 10==20.9489 = the arc of the given seg- 
ment. ; 

Then 20.9439 


— 8.6602 
ABD. 
Or, taking the same example, having found the length of the 
arc ADB =20.9439, we have by Rule II. 


20.9439 aa +) 
—17.3205 5+17.8205-+$==61.418 = the area the same 


as before. 


5, = 61.418 = the area of the segment 


190 MENSURATION 


Scholium. If two of the following parts, viz: the chord, 
versed sine, and radius are given, the other may be found by 
the formule (in mensuration Hl..Geom. Prop. XIII.) 


Scholium 2. The triangle ACB is = the triangle AFC. 
EBxCE=1CBx AS. 
That is, the product of the sine of an arc X its cosine = } 
sine of twice. the arc X radius. 


CASE V.. 
When the chord and radius only are given. 


Ruxe.—Divide half the chord by the radius, and the quo- 
tient will be the sine of half the arc of the segment ; find the 
arc corresponding thereto in the table of natural sines, in de- 
grees, and multiply it by 2; then find its length as in Rule 3, 
and multiply it by the radius. 

Take the versed sine — ED—CD—vy(CA—AE.) Then 
proceed as in Rule 2 to find the area of the segment. 


Ex. Taking the same example as in the last rule, having: 
found the arc = 20.9439, we have 
ED — 10 — J/ (100 — 69,899905404) = 5. 


Hence, 20.9439) . , hs 
17.3205 UP tn pee es = 61.418. 


Examples for Practice. 


Ex. 1. What is the area of a segment whose arc is 3.14159 
and whose sine is .87785, the radius being 10 7 


Ex. 2. Required the area of the segment whose chord is 


== 12, the radius — 10. Ans. 16.35. 
Ex. 3. What is the area of the segment whose height is 2, 
the chord being 20 ? Ans. 26,8804, 


Ex. 4. What is the area of a segment of a circle whose 
arc is 110° the radius being 1? 

Ex. 5. Required the area of the segment whose height is 5, 
the diameter being 8. Ans. 33.0486. 


PROBLEM If. 


To find the area of a circular zone AKDB, or the space in- 
cluded between two parallel chords AB, ED, and two arcs 
AE, BD. 


Rute. Multiply the two ares AE, BD, of the zone by } the 
radius, to which add the sum of the products of the sines of the 


OF SUPERFICIES. 191 


external segments, with ' the radius, if on different sides of the 
centre, or add the difference of those products, if on the same 
side. (Prop. XXIII Schol. Formula 3, B. IV.) 


A = 3(rv'—rs')—}(rz—rs,) 


L 
Ex. What is the area of a zone ABDE, 
the sum of whose arcs AE+BD is=6,28318 , xe 
the sine of the arc ALB=2,57178, and the 
sine of the arc EFD = 2,62386, the radius 
being = 3? A 
vols tee 


F 


6,28314 
-++ 262386 3 = 17,31817 the area required. 
+257178 


Scholium. The same rule will apply 
to any portion ABEF of the circle includ- 
ed between two chords, that are not par- 
allel. Hence, if the sum of the axes AF, 
AE in this figure, is equal to the sum of 
AE, BD in the last, and if the arcs AIB, 
FLE in this are respectively = ALB, 
EFD in that, then the portion AFEB in 
this will be = the zone AEDB in that. 


Ez. 2. What isthe areaof azone AGHB whose two chordsare 
on the same side of the centre, the sum of the arcs of the zone 
being 2,09436, the sines of the arcs on each side being 2,59806, 
and — 2,95440, the radius being = 3? 

2,09436 
+2,59806 X 2 = 2,60703 the area required. 

—2,95440 


Nore. Other rules have been given for finding the areas 
of circular segments and zones in Mensuration, El. Geom ; 
formule are there given for finding such data as are required, 
for the elements of the area. 


Examples for Practice. 


Ex. 1. Required the area of the 
zone ABEDHGH, the greater chord 
AB = 186 feet, the less chord HD 
= 68 feet, and the distance LP = 
248 feet. 

Ans. 55655.1965159 sq. feet. 


« 


192 MENSURATION 


Ex. 2. Suppose the zone to have its parallel chords equally 
distant from the centre of the circle O, each chord AB and 
HD = 12.49 feet and their distance LP = 10 feet ; required 
the area of the zone. Ans. 148.86672 sq. feet. 

Ex. 3. Supposing the circular zone ABEDHGA, having its 
greater parallel chord = 40 yards, being equal to the diame- 
ter of the circle, the less chord = 20 yards, and their distance 
LP = 17.310508 yards ; required the area of the zone. 

Ans. 592.08244 sq. yards. 

dix. 4. Required the area of the zone ABEDHGA, the pa- 
rallel chords AB, and HD, being 16 feet and 12 feet, and their 
distance HP = 14 feet. Ans. 253.0792 sq. feet. 

fix. 5. The zone, whose parallel chords AB = 40, HD = 
30, and the breadth = 35 ; required the area of the zone. 

Ans. 1581.745. 

Ex. 6. Suppose the two parallel chords AB and HD = 80 
feet and 60 feet, and the perpendicular distance from each 
other = 70 feet; it is required to find the distance of the 
greater chord AB from the centre atO ; and also to find the 
radius of the circle. 

The distance OP. .... ..... = 30 feet = 1st Ans. 
The radius of the circle OF = 50 feet = 2nd Ans. 

Kix. 7. Required the area of the zone ABEDHGA, whose 
arcs AGH, BED are together = 160°, the arc AOB being 
110°, and the radius of the circle 10 feet. 

Fix. 8. Required the area of the portion ABEF, included 
between the two oblique chords AB, FE, (see diagram to 
Scholium above,) whose arc AF= 60°, BE=80°, and AIB = 
110°, the radius being 20. 


PROBLEM III. 
To find the circumference of a circle approximately. 
CASE I. 
When the radius sine, and cosine of any small arc is given. 


Rue 1. Divide 11 times the product of the radius and sine 
by the sum of the radius, and half the cosine, which will give 
the length of the arc, multiply this by the number of such are 
in the whole circumference, and the product is the circumfer- 
ence of the circle ; the accuracy of which depends in the small- 
ness of the arc. 

Let s — the sine, 

c = the cosine, 


OF SUPERFICIES. 193 


and ’ — the arc corresponding to these functions, 
* — being the semi circumference, and r the radius of 
the circle = 1. | 


v ‘ 
«= ih being the number of parts each = «’ that the 


_ semi-circle is supposed to be divided into. Then (Prop. IX, B. 
IV. Formula 1.) 


wae 2rs 
r+ic 
T 
Letia? ==. 
“™ 20000 
Then s = ,000157079632033 
c= ,999999987462994 
and rs = ,000235619448050 
r+le = 1,499999993831497 
3rs 
Hence —-——= _,000157079632676 
» rT +46 


and, 000157079632676 xX 20000 = 3,1415926535,2 = the 
circumference of the circle, whose radius is 1, which is true to 
the last figure, which should be 8 instead of 2. 


Scholium. Other formula may be found for determining the 
circumference of a circle at Prop. IX B. XIV, viz., formule 2 
and 3, to which the student is referred ; at which place, will 
also be found some important trigonometrical formule for 
finding the value of the sines and cosines, and other functions 
of the circle. 


Rute 2. Divide 6 times the product of the radius and 
versed sine of a small arc, by the sum of the sine + 4 times 
the sine of half the arc, which will give the length of the arc, 
which, multiplied by 2, the number of times this arc is con- 
tained in the circle, gives the circumference. 


6A’ 
vow ie . XVII, 
= .5 +b (Prop I,) which may be expressed 


6r.versin.«’ — 6rv 
sin. t + 4sin.in! — s+4s' 
Ex. Having the sine and cosine of anarc=to ;1, part ofa 


quadrant= ,0157073173118 = s 
and ,9998766324816 = c; it is required from these data_to 
find the arc 7’ the radius being 1. 
The versed ‘sine = 1 — ,9998766324816 
= ,0001233675183 = v 
Hence, 67v = 0007402051098 
And by Trigonometry we shall find s’ = ,0078539008887. 
Hence, s + 4s’ = ,04771229208666. 


194 © MENSURATION 


: orv 
And 0007402051098 — 0471229208666 = 3-4! 


= ,015707963267,5 = s' which is true to the last figure, 
which should be 9 instead of 5. | 


6A’ 

peaeay also be put under the — 
6rs " 6rs 

Cos. «/-+4 Cos. x! e-F4e' 


iz. 2. Let it be required to find the value of 2’ to twenty 
decimal places ; for this purpose let x’ be an are of yytz— part of 
a quadrant, or 1 minute, according to the French centesimal 
divisions of the circle. The sine of this are to 21 decimal 
places, according to Legendre’s Trigonometry, is 
= .000157079632033525563=s 
and its cosine=.999999987662994524005 =c 


piesa 
And by the trigonometrical formula cos. 6= <4 1 + cos. ° 


Scholium. The expression 


form 


2 
we have cos. de! = .999999996915748676195. 
Hence we have 67s=.000157079632033525563, 
and 6+-4c’ =,5999999975325989228785. 
6rs : ; 
Therefore,- x’ = 0001570796326794896,5, which is 


c+4e’ A 
true to the-last figure, which should be 6 instead of 5. 


Scholium. Since there is no limit tothe smallness of the are, 
which may be taken, and since its sine and cosine may be cal- 
culated to any number of decimal places whatever, it there- 
fore follows, that there is no limit to the accuracy with which 
the circle’s circumference may be calculated by this method. 

The circumference of the circle, as found by M. DeLagney, 
to 128 decimal places, by a method furnished by the calculus, 
is as follows 

The circumference of a circle whose diameter is 1, is 
3.1415926535897932384626433832795028841971693993751 

058209749445923078 1640628620899862803482534211706 

798214808651 32723066470938446 + or 7—. ’ 


The series has more recently been extended to 154 decimal 
places. We might proceed by this method to verify the re- 
sults obtained by the calculus, and extend the number of 
decimals much farther, were it worth the labor ;- but since 
we have the result already, extended beyond what is prac- 
tically useful, the labor may be reserved for those who. have 
leasure and inclination to pursue it. It may be shown that, 
however far, the circumference should be developed in 
terms of the diameter, the expression would never terminate, 


OF SUPERFICIES. 195 


or in other words, the circumference and diameter of a 
circle arc incommensurable in terms of each other, See notes. 


_ PROBLEM IV. 
1. To describe an Ellipse. 


Let TR be the major axis, CO the Nor 
minor axis, and c the centre. With 
the radius T’c and centre C, de- 
scribe an arc cutting TR in the 
points F, f; which are called the 
two foci of the ellipse. 


“D> 

Assume any point P in the major axis ; then with the radii 
PT, PR, and the centres F f, describe two arcs intersecting 
in I; which will be a point in the curve of the ellipse. 

And thus, by assuming a number of points P in the major- 
axis, there will be found as many points in the curve as you 
please. Then with a steady hand, draw,the curve through 
all these points. 

Otherwise with a Thread. 


Take a thread of the length of 
the axis-major AB, and fasten its 
ends with two points in the foci, 
SH. Then stretch the thread, and 
it will reach to P in the curve: and 
by moving a pencil round within 
the thread, keeping it always 
stretched, it will trace out the el- 
lipse. 

There are various instruments used for the construction of 
this and the other conic sections. But we have not room, 
consistantly with our plan, to describe them here. 


PROBLEM V. 


In an ellipse having either three of the following parts given, 
viz., the major or minor-azxis, the ordinate, or abscissa, to find 
the fourth. 

CASE, Ix 


To find the ordinate. 


When the major axis, the minor axis and abscissa, are given. 


Rute. As the major-axis is to the minor-axis, so is the square 
root of the product of the two abscisse to the ordinate. 


106 MENSURATION 
Fix. 1. In the ellipse of ABHD  _B 


the major-axis AH = 70, the minor- 
axis BD = 50, and the two abscis- 
se AS = 14, HS = 56, itis requir- 
ed to find the length of the ordinate 4 Hi 


AH: BD:: y(AS x HS): LS, 
viz, 70: 50:: v(14 X 56) : 20, the length or the ordinate 
required. 


Ex. 2. If the major, and minor-axes, of an ellipse are 80 and 
60, the abscissee AS = 16, what is the length of the ordinate ? 
Ans. 24. 


CASE Il. 
To find the two abscissa. 
When the major, and minor axes, and ordinate are given. 


Rute. As the axis-minor is to the axis-major,so is the square 
root of the difference of the squares of the semi-minor axis 
and ordinate, to the distance between the ordinate and centre ; 
which distance, added to and subtracted from, the semi-axis 
major will give the two abscissz. 


Ex. 1. The major-axis AH = 10, its conjugate BD = 50, 
ang the ordinate LS = 20; required the two abscisse AS, 

S. 

BD: AH:: y( (£BD)’—LS’) : CS, viz., 

50: 70:: ¥ (25*—20’) : 21, the distance from the centre 
to the ordinate. 

Hence, 2AK SC "(70 2) 235 220 = "56 
and 14 = AS, HS, the two abscisse. 


2. What are the two abscisse AS, HS, the ordinate LS = 
24, and axes AH BD = 80 and 60? 


Answer 16 and 64. 


3. The major-axis AH = 36, its conjugate BD = 24, and 
ordinate LS = 8 ; required two abscisse HS, AS. 


Answer 18 + 3/2= 18 + 4,2426408 = 22,2426408 and 
13,7573592. 


OF SUPERFICIES. 197 
CASE UL. 
To find the major azis. | 
When the minor axis, ordinate, and abscisses, are given. 


Rute. From the square of half the minor axis, subtract the 
square of the ordinate; then extract the square root of the 
remainder. Next add this root to the semi minor axis, if the 
less abscissa be given, but subtract it if the greater abscissa 
is given, reserving the sum or difference. Then say as 
the square of the ordinate, is to the rectangle of the 
abscissa and minor axis, so is the reserved sum or difference 
to the major axis. 


Ex. 1. In the ellipse ABHD, there are given the minor axis 
BD = 50, the ordinate LS = 20, and the less abscissa AD = 
14 ; required the major axis AH. 


First. /{ (:BD)’ — LS’) = v(16? — 20’) = v225 = v 
(5° X 3°) = 5 X 3 = 15, the square root of the difference of 
the semi-conjugate axis, and the ordinate. 

Then }3BD + 12 = 25 + 15 = 40, the sum. 


Secondly. LS’: BD x HS: : 40: AH, viz., 207: 50 x 14:: 
40:70 = AH, the major axis required. 


Ex. 2. If the minor axis BD = 40, the ordinate CS = 16, 
and the less abscissa AS = 36; what is the length of the ma- 
jor axis AH. Ans. 180. 

CASE IV. 


To find the minor azis. 
When the major axis, ordinate, and absbiss@, are given. 


Rute. As the square root of the product of the two abscis- 
sz is to the ordinate, so is the major axis to the minor axis, 

Ex. 1. The major axis AH = 180, the ordinate HS = 16, 
and the greater abscissa HS = 144; required the length of 
the conjugate axis AD. 


Here AH — AS = 180 — 144 = 36 = HS, the less ab- 
scissa. 
Then /(ASXBS) : LS: : AH: BD, viz., y(144X 36) : 16 
: ; 180 : 40, the conjugate axis AH. 
Ex. 2. The major axis AB = 70, the ordinate LS = 20, 
and the abscissa HS = 14; required the ipa BD. 
ns. 50. 


198 MENSURATION 


' 
PROBLEM VI. 
To find the area of an ellipse. 
CASE I. ¢ 
When the major and minor axes are given. 


Ruts. Multiply the product of the semi axes by 7= 3,14159 
or the circumference of a circle whose diameter is 1, and this 
product will be the area. : 


Fiz. 1. Required the area of an 
ellipse ABLD, whose axes are AL 
= "70, ahd BD = 50. 
PAL X IBD Xx*= 10 X 50 X 
3,14150 = 2748. 9, the area of the ,4 = 
ellipse required. 


Ex. 2. What is the area of the ellipse whose major axis 1s 
23, and the minor axis = 18 ? Ans. 339.2928. 


Fix. 3. The major and minor axes being 61,6, and 44 re- 
spectively, required the area of the ellipse. 
Ans. 2128.7481,6. 


Ex. 4. What is the area of an ellipse, whose axes are 25 
and 19 ? Ans. 373,06381. 


iz. 5. What is the area of an ellipse whose axes are 23, 
and 17 respectively ? Ans. 307,09042. 


Scholium. If there be two or more concentric ellipses 
FGHK, fghk, the area of the inner one subtracted from that 
of the outer one, will be the area of the elliptical ring included 
between them. 


Hence, also as for a circular ring (Mensuration Hl. Geom.) 
so with the elliptical ring, its area is equal to the difference of 
the rectangles of the semi axes of the inner one and outer one, 
multiplied by v = 3,14159. 

Let the ellipse be taken, whose axes are 25 and 19 ; 23 and 
17, in the last two examples ; 

25 19 23 17, . 
and we have Cora xX Pe cera isy x ares w= (12,5 x 9,5— 
11,5 X 8,5) « = (118,75 — 97,75) 7 = Q2le* = 65,97339 = 
he area of the ring Ff, Gg, Hh, KA. 


"+ 


OF SUPERFICIES. __ 199 


Let the results be taken from 
the two examples referred to, 
and the area of the ring will be 
found to agree with this, viz., the 
area of the outer ellipse is there 
found - -  «=373,06381 
The area of the inner 
one- = - =807,09042 
The difference is =65,97339 
the same as found above 


CASE Il. 
When any two conjugate diameters are given. 


Roxe. Multiply continually together any two semi conjugate 
diameters, the sine of their included angle, and x. (Prop. IV, 
Cor. 5, B. I.) 


Ez. The two conjugate diameters yan wy ty 
AB, FG, of the ellipse ADFBGEA 


being 32 and 28, and their included dif_p< A r Wey B 


angle 77° 344’; required its area. 


The sine of 77° 34’f is 9765625 ; 
therefore, 9765625 xX 16 xX 14x 
3,14159 = 687.225 = the area. 


PROBLEM VII. 


To find the area of the segment of an ellipse, cut off by an 


ordinate to any diameter. 
CASE I. 


When the ordinate is perpendicular to either of the principal 
axes. 


Rute. Find the corresponding segment of a circle of the 
same height, described on the same axis, to which the cutting 
line or base of the segment is an ordinate. 

Then, as this axis is to its conjugate, so is the circular seg- 
ment to the elliptical segment. 

Or find the area of a circular segment, whose versed sine or 
height is equal to the quotient of the height of the elliptic seg- 


y* 


200 MENSURATION 


ment divided by its axis. Then multiply continually together, 
this segment and the two axes of the ellipse, for the area of 
the segment required. — 


Ex. What is the area of an 
elliptic segment JLA, cut off 
by the line IL, parallel to, 
and at the distance of 7 
from the minor axis EF’, the 


G 
axes being 35 and 25 ? ne 
H 


17: — 74 = 10 the height Ae 
of the segment. 


Then 2./(Ae X Be) = GH the corresponding ordinate or 
chord to a segment of the circumscribing circle=2./(10X 25) 
= 15,8113883 X 2 =31,6227766. 

Let the semi chord Ge be divided by the semi diameter, 
and we shall have the corresponding sine of the arc GA of a 
circle, whose radius is 1 = 15,8113888 ~~» 17,5 = .903508) 
corresponding to which, is the arc of 64° 373’; hence the 
arc GAH = 64° 37)’ X 2= 129° 15’. 

Then AB X * = 35 X 3,14159 = the cireumference of the 
circle ACBD = 109, 95565. 

And 360° : 129° 15’ : : 109,95565: 39,4771 = the areGAH 
ofthe segment ; and by Problem I, Case IJ, 

(39,4771 — 31,6227) Sr + 31,6227 X 10 + 2 = 

17,8544 X 8,75 + 31,6227 X 5 = 68,726 + 158,1135 = 

226,8395 = the area of the circular segment GHA. 

Then 35 : 25, or7 : 5: : 226,8395 : 162, 171 = the area of 

. the elliptic segment ILA. 


Scholium 1. If the area of the segment ILFBE had been 
required, the circular arc GCBDH should have been taken 
instead of the arc GAH. 

2. If the segment Io, whose base is parailel to the major 
axis, is required, it may in like manner be found from its re- 
lation to the segment POE of the inscribed circle. 


OF SUPERFICIES. 201 
CASE Il. 
When the base of the elliptic segment is oblique to the axes. 


Rute. Divide the abscissa Pv by its diameter Pp, and find a 
circular segment whose versed sine or height is the quotient. 
Then multiply continually together the area thus found, and 
the two axes, for the elliptic seyrment. Or multiply continually 
together the circular segment, the diameter Pp, to which the 
base of the segment is a double ordinate, its conjugate diame- 
ter Dd, and the sine of their included angle, for the area of the 
elliptic segment. 


x. The principal axes of an ellipse 
being 35 and 25, it is required to find 
the area of a segment QgP, whose 
base Qq is an ordinate to the diameter 
Pp, whose length is 33, it being divid- 
ed by the ordinate into the two ab- 
scisse Pv = 7, and pv = 26. 


FA + AB = 7 ~ 33 = .2121,7,= the versed sine or height 
of the segment. 
The area of a segment corresponding to this height in a cir- 
_ cle, whose diameter is 1, is .12162866. 
Hence, .12162869 X 25 X 35 = 106.4251 = the segment 


QgP. 


PROBLEM VIII. 


To find the circumference of an ellipse. 


First, find the area of an elliptic ring included between an 
interior and exterior concentric ellipse, whose axes are sever- 
ally the axes of the given ellipse + n, and the same axes 
— nj; then divide this area by the average distance between 
the exterior and interior curves, and the quotient will be the 
circumference of the given ellipse. (Prop. VIII, B. IV.) 


Ex. Let it berequired to find the circumference AEBDA of 
an ellipse, whose axes AB, ED are 24 and 18. (See diagram 
to Scholium, Prob. VI.) 

Let us assume two other exterior and interior concentric 
ellipses, whose axes FH, GK are = AB + n, and ED + 2; 
and fh, gk, are = AB —n and ED — 2; and if n = 2, we 
shall have the area of the ring, as found in examples under 


+ 14 


202 MENSURATION 


Scholium Prob. VI=65,97339. Let this be divided by the 


. : Ae 9899+! 
average distance as found in Proposition VII, B. v= 


and we have 66,3115 for the elliptical circumference AEBDA. 


Scholium 1. If great accuracy is not required, the following 
approximating rules from Hutton’s Mensuration may be used. 


Ruue 1. Multiply the sum of the semi axes by =, or 3,1416, 
and the product will be the circumference nearly. 


Ex. Required the circumference of an ellipse, whose axes 
are 24 and 18. : 
(12 + 9) + 314159 = 21 X 3,14159 = 65,9735 equal the cir- 
cumference nearly. 


Or Rue 2. Multiply the square root of half the sum of 
the squares of the two axes by 7, and the product will be 
nearly = the circumference. 


Ex. Taking the same example as before, we have 


24 + 17° : 
: 5) xX 3,14159 = 66,6483 = the circumference 


nearly. 


It will be observed by comparing the last two results, 
that the former one is nearly as much in defect, as the latter 
is in excess ; hence, if we take half the sum of the two we 
shall have the circumference of the ellipse more accurately. 


Thus + pes + 2 = 66,3084, which is very neaf 
the truth. 


PROBLEM IX. 


To construct a parabola ; having given any ordinate PQ to 
the axis, and abscissa VP. 


First, find the focus F thus ; 
bisect PQ in A; draw AV, and 
AB perpendicular to it; take VF 
= PB, and F will be the focus. 


Arithmetically. Divide the < 
square of the ordinate by four 
times the abscissa, and the quo- B 
tient will be focal distance VF. * 


OF SUPERFICIES. 203 


Then, in the axis, produced without the vertex V, take VC 
= VF; draw several double ordinates SRS ; then with the 
radii CR, and the centre F’, describe arcs cutting the corres- 
ponding ordinates in the points S. — 

Draw the curve through all the points of intersection, and 
it will be the parabola required. 


PROBLEM X. 


Of any abscissa X, its ordinate y, and latus rectum, or para- 
meter p; having two given, to find the third. 


+ 


CASE 1. 
To find the latus rectum. 


Divide the square of the ordinate by its abscissa, and the 
quotient will be the latus rectum. 
Or, take a third proportional to the abscissa and ordinate, 
for the latus rectum. 
That is, p = y’? + =. 


EXAMPLE. 


If the abscissa be 9, and its ordinate 6 
Then 6 X 6 + 9 = 36 + 9 = 4 = the latus rectum. 


CASE IL. 
To find the abscissa. 


Divide the square of the ordinate by the latus rectum, and 
the quotient will be the abscissa. 
That is,z =y'? > p. 


EXAMPLE. 


If the ordinate be 6, and the latus rectum 4. 
Then 6 X 6 +4=367+4=9= the abscissa. 


CASE Ill. 
To find the ordinate. 


Multiply the latus rectum by the abscissa, and the square 
root of the product will be the ordinate. 
That isy = pz. 


* 


204 MENSURATION 
EXAMPLE. 


The absciss being 9, and the latus rectum 4. 
Then v(9 X 4) = 36 = 6 = the ordinate. 


PROBLEM XI. 


Of any two abscise A, B, taken upon the same diameter, and 
their two ordinates a, b ; having any three given, to find the 
fourth. 


The abscissz are to one another as the squares of their or- 
dinates. That is, as any one abscissa is to the square of its 
ordinate, so is any other abseissa to the square of its ordinate 3; 
and conversely. Or, as the root of one abscissa is to its ordi- 
nate, so is the root of any other abscissa, to its ordinate. 


B «avAB 
SALnLB >: was Ovigrm_g ae 


a VA::b:byA _bvAB 
Pisce 4 
( 


Hence 


Bas 
3 Ab’ 
And : a =— so B 
2 . B od 


Ez. 1. If an abscissa = 9 correspond to an ordinate = 6, 
required the ordinate whose abscissa is 16. 
Here V9: Y16::6:6 X 4 + 3 = the ordinate. 


Ex. 2. Required the abscissa corresponding to the ordinate 
6, the ordinate belonging to the abscissa 16 ole} S. 
Here 8’: 6’: : 16: 9 = the abscissa. 


PROBLEM XII. 


To find approximately the length of any arc of a parabola, cut 
off by an ordinate to the axis. 


| When the abscissa and ordinate are given. 


Rue. To the square of the ordinate add four thirds of the 
square of the abscissa, and twice the Pe root of this sum 
pa be the tea of the curve, nearly.* 


* See Hutton’s Mensuratien. 


# 


OF SUPERFICIES. 205 


Fx. The abscissa VH = 2, and the ordinate AB = 6, re- 
quired the length of the curve EVE. 


Here 2, (AB? + 4VB’) = 2v [6+ 
Qi + 3)) = 2v-34 = 2 V3l X 3= 
3793 = 9.6436508 X 3 = 12.8582, A 
the length of the curve AVC, nearly. 


Examples for Practice. 


Ex. 1. What is the length of the parabolic curve AVG, 
whose abscissa VB = 2, and the ordinate AB = 8 ? 
Ans. 17.4356. 


Ex. 2. Required the length of the parabolic curve DAVCF, 
when the abscissa VE = 16, and the ordinate DE = 12. 
Ans. 42.142615. 


Ex. 3. Required the length of the parabolic curve DAVCF, 
when the abscissa VE = 8, and the ordinate DE = 16. 
Ans. 36.951. 


PROBLEM XIII. 


To find the area of a parabola, when the base and height 


are given. 


Ruts. Multiply the base by the height, and two-thirds of 
the product will be the area. 


Ex. Required the area of the parabola AVCA, the abscissa 
VB = 2, and the base, or ordinate, AC = 12. 

Here 2 (AC x VB) = 2(12 X 2) = 16, the area of the 
parabola AVCA required. 


Examples for Practice. 


Ex. 1, What is the area of a parabola DAVCFD, whose 
abscissa VE = 10, and the double ordinate DF = 16. 


Ans. 1062. 


& 


206 MENSURATION | 


Fix. 2. Required the area of a parabola DAVCFD, whose 
base or ordinate DF — 15, and the abscissa VE — 22? 
Ans. 220. 


Ez. 3. What is the area of a parabola AVCA, the base or 
ordinate AC = 20, and the height or abscissa VB = 6. 
Ans. 80. 


PROBLEM XIV. 


To find the area of parabolic frustum, or zone of a parabola, or 
of the space included between two parallel ordinates. 


The two ordinates, and their distance being given. 


Rute. To the sum of the squares of the two ordinates, add 
their product, divide the result by the sum of the two ordinates, 
the quotient multiplied by two-thirds of the altitude of the 
frustum, will give the area. 


Hix. Required the area of the parabolic frustum ACFDA, 
the two parallel ordinates DF’, and AC = 10, and 6, and the 
distance BE = 4. 


(DF’ + AC’) + (DF x = pails 
fet onpE PAG o> x3 BB 
by vasa ore), 2 1386+60 8 
Tt Vie (HIOGE Os 2, surat, waameae a TO Minates 
196 8 98 . 
=syo*3=5 = 322, the area of the frustum ACFDA. 


Examples for Practice. 


fix. 1. What is the area of the parabolic frustum ACFDA, 
whose two ordinates DF and AC — 10 and 6, and the dis- 
tance BE = 3? Ans, 241, 


Ex. 2. The greater end of the frustum DF = 30, the less 
end AC = 20, and their distance BE = 15; required the 
area. Ans. 380. 


Ex. 3. The greater end of the frustum DF — 20, the less 
end AC = 10, and their distance BE = 12. Ans. 1862. 


OF SUPERFICIES. 207 


OF THE HYPERBOLA, 
PROBLEM XV. 


To construct or describe a hyperbola. 


Let o be the centre of the hy- 
perbola, or the middle of the 
transverse AB; and BC per- 
pendicular to AB, and equal to 
half the conjugate. 

With the centre o, and radius 
Co, describe the circle, meeting 
AB produced in F and f, which 
a the two foci of the hyper- 

ola. 


Then assuming several points vv, &c., in the transverse 
produced, with the radii Av, Bv, and centres f, F, describe arcs 
intersecting in the several points g, g, &c., through which 
points draw the hyperbolic curve. 

If straight lines oM, oN, be drawn from the point 0, the mid- 
dle of the transverse diameter, through C, and D, the extrem- 
ities of the conjugate, they will be the asymptotes of the hy- 
perbola, the property of which is to approach continually to 
the curve, but not to meet it, until they be infinitely produced. 


PROBLEM. XVI. 


In an hyperbola to find the'transverse axis or conjugate axis, or 
ordinate or abscissa. 


CASE I. 
To find the ordinate. 


When the transverse axis, conjugate axis, and the abscissa 
are given. 


Rute. As the transverse axis is to the conjugate axis, so is 
the square root of the product of the two abscisse to the 
ordinate. 

Nors. In the hyperbola, the less abscissa added to the axis? 
gives the greater abscissa. 

Ez. If the transverse axis AB = 24, the conjugate axis CD 
= 21, and the less abscissa BH = 8, what is the length of the 
0 rresponding PH. 


208 MENSURATION 


Here AB: CD:: y[(AB+ BH) x BH]:: PH, viz. 24: 
21:: /[(24+ 8) x 8]: 14, the length of the corresponding 
ordinate PH, required. 


Examples for Practice. 


Ex. 1. The transverse axis AB = 60, the conjugate axis 
CD = 36, and the less abscissa BH — 20, required the cor- 
responding ordinate PH. Ans. 24. 


Ez. 2. The transverse diameter AB = 50, the conjugate 
diameter CD = 40, and the greater abscissa AH = 64; re- 
quired the ordinate PH. Ans. 32/14. 


Ex. 3. Required the length of the ordinate MK, whose 
transverse axis AB = 609, the conjugate axis CD = 588, and 
the less abscissa BK = 116. Ans. 280. 


CASE Il. 
To find the two abscisse. 


When the transverse axis, the conjugate axis, and the ordinate, 
are given. 


Ruts. As the conjugate axis is to the transverse axis, so is 
the square root of the sum of the squares of the ordinate and 
semi-conjugate to the distance between the ordinate and cen- 
tre, or half the sum of the abscisse. Then will the sum of this 
distance and the semi-transverse be the greater abscissa, and 
their difference the less. 


Ex. The transverse axis AB — 24, the conjugate axis CD 
= 21, and the ordinate PH = 14; required the two abscisse 
AH, and BH. 

Here CD: AB:: y[PH? + (}CD)’] : HO, viz. 

21: 24:: v(142 + 10.5%): 20. 

Then the two abscissas AH and BH =HO + 1AB= 20 + 

12 — 32 and 8. 


Examples for Practice. 


Hix. 1. The transverse axis AB — 60, the conjugate axis 
CD — 36, required the two abscisse AH, and BH, corres- 
ponding to the ordinate PH = 24. 

Ans. AH = 80, and BH = 20. 

Ex. 2. The transverse axis AB = 120, the conjugate axis CD 
= 72, and the ordinate MK = 48, required the two abscisse 
AK and BK. Ans. AK = 160,and BK = 40. 


OF SUPERFICIES. " "209 
PROBLEM XVII. 


To find the length of any arc of an hyperbola approximately 
beginning at the verlex. 


When the transverse and conjugate axis, the ordinate, and 
abscissa, are given. 


Ruts. First. Add 21 times the the square of the conjugate to 
19 times the square of the transverse, and multiply this sum by 
the abscissa; to this product add 15 times the transverse, 
multiplied by the-square of the conjugate, and call this quan- 
tity the dividend. 

Secondly. Add 21 times the square of the conjugate to 9 
times the square of the transverse, and multiply this sum by 
the abscissa ; to ‘this product add 15. times the transverse, 
multiplied by the square of the conjugate, and call this quan- 
tity the divisor. 

Thirdly. Then divide the dividend by the divisor, and mul- 
tiply the quotient by the ordinate for half the length of the 
curve, or multiply the quotient by twice the ordinate for the 
length of the whole curve, nearly.* 


Ex. 1. Required the length of the hyperbolic curve PLBRG 
to.the abscissa BH=2.1637, and the ordinate PH=10 ; the 
two axes AB and CD=80 and 60. 


Here (2ICD'H19AB’)x BH +(15ABXCD") oy 
Mere (91CD'+ 9AB') xX BH+(15ABXOD*)*~ 
__[(21 x 60’) + (19 X 80") ] x 2.1637-+(15 X 80 X 60") 
~ [(21X 60°) +( 9X80") |X 2.1637 + (15 X 80 X 60) 
__ (75600-+121600) x 2.1637+4320000 
~ (75600-+ 57600) X2.1637-+4320000 *7° 
426681.64+4320000 4746681.64 
= "288204.84+4320000 * 7° —4608204.84 * 20 = 1.03005 x 20 
=20.601, the length of the whole curve PLBRG required. 


*2X10 


—_ 


* Hutton’s Mensuration. 


210 MENSURATION 


Ex. 2 Required the length of the hyperbolic are PLBRG, 
the abscissa BH = 20, the ordinate PH = 24, and the two 
axes AB and CD =60 and 36. Ans. 62.652. 


PROBLEM XVIII. 
To find the area of an hyperbola. 


When the transverse axis, conjugate axis, and the abseissa, are 
td 
given. 


Rute. To the product of the transverse axis and ab- 
scissa, add } of the square of the abscissa, and multiply the 
square root of the sum by 21 ; to this product add 4 times the 
square root of the product of the transverse axis and abscissa ; ;- 
then multiply this sum by 4 times the product of the conjugate 
axis and abscissa, and divide this last product by 75 times the 
transverse axis, the quotient will give the area of the hyper- 
bola, nearly. * 


Ex. 1. Required the area of the hyperbola PBGP, whose ab- 
scissa BH = 10, the transverse and conjugate axis AB and 
CD = 30 and 18. 

{21 v[ (ABx BH)+BH’]+4v(ABXRH)} x4CDx BH 
ae 75 AB 
_ f21 v[(80x 10) +(§ x 10*)]+-4 v(30x 10)}x4x 18x10 
ie 75 X 30 
1 J/371¢+4/300)X8 (21 J 2529 +-40/3)X8__ 
ea DS ak Wi tithe ik veae he 


8 10 
siz X 5 \/ (267) +40 8) 55 X (30 v182+-40 3) = 
8x10 

a x (8182448) = = 18, (40.4722128-+-6.9282082)= 


151.681328, the area of te hyperbola PBGP required. 


Ex. 2. What is the area of the hyperbola MBNM, the ab- 
scissa BK = 25, DP transverse and conjugate axis AB and 
CD = 50 and 30? Ans. 805.090844. 

PROBLEM XIX. 


To find the area of any mizxtilineal figure by means of equidis- 
tant ordinates, terminated by a curve on one side, and aright 
line as a base on the other. 


Rute. To the sum of the first and last ordinates add 4 times. 
the sum of all the evem ordinates, and twice the sum of all the 


* Hutton’s Mensuration. 


OF SUPERFICIES. 211 


odd ordinates, rejecting the first and last; and 1 of this result, 
multiplied by the common distance of the ordinates, will give 
the area, very nearly. Prop. XV, B. IV. 


Scholium. This rule is absolutely true for a parabola, if the 
ordinates are parallel to its axis. And if the distances be- 
tween the ordinates is small, it is approximately true for any 
other curve. 


Ex. 1. Required the area of an irregular figure, bounded on 
one side by a curve line at five equidistant ordinates, the 
breadths being AD=8.2, mp = 7.4 nq=9.2, or = 10.2, BC = 
8.6; the length of the base AB = 39, and the common dis- 
tance of the ordinates Am, mo, no, oB, each = 9.75. 

Here + [((AD+BC)+4 (mp p q 
+ or) + 2ng}) X° R75 = | | c 
11(8.2 + 8.6) +4(7.4 + 10.2) 

+ (9.2X2)] X 9.75 = 4(16.8 
+ 70.4-+ 18.4) X9.75=(105.6 5-7 EN OTT 
+3) X 9.75=343.2, the area of the space ADCBA required. 


fix. 2. Required the area of an irregular space ADqnA 

bounded on one side by acurve line, and divided by three 

equidistant ordinates perpendicular to the base An, the or- 

dinates being AD=8, mp=6, and nq=10, the length of the 

base An=14, and the common distance Am, mn, each equal 7. 
Ans. 98, 


Ex. 3. The abscissa of a parabola being 2, and the base 
or ordinate 12, required the area of the parabola. 

Here, by taking three ordinates, of which the first and last 
are each nothing, the middle one being the abscissa=2, and 
the common distance=6 ; hence the area of the parabola=16 
= Ans. 


212 MENSURATION. 


MENSURATION OF SOLIDS. 


PROBLEM IT. 


To find the solidity of a sphere, spheroid, a spherical or an 
elliptical revolord. 


Rute.—Multiply a central conjugate section by the vertical 
axis, and take two-thirds of the product for the solidity. 

Ex. 1. What is the solidity of a sphere, whose Zs 
diameter is 10 feet ? if «o 

31.4159 xX $=78.5397 = toa central sec- 4 \iC <A’ 
tion; hence, 78.5397 10X2 = 523.931 cubic “ jj 
feet the solidity. 

Kix. 2. What is the solidity of a prolate 
spheroid, ACBD, whose vertical or fixed 
axis, AB, is 10, and its revolving axis, CD, 
is 5! 

3.14159 X5X 11 = 19,63494 = a cen- 
tral conjugate section. 

Hence, 19.63494 x 10 X 2=130.9329, the solidity required. 

Kix 3. Required the solidity ~ 

of a spherical hexagonal re- 
voloid, BCGEDF, whose ver- 
tical axis is ten feet. 
- By referring to the table of 
Polygons, (Mensuration El. 
Geom.) we find the area of a 
hexagon, circumscribed about 
a circle whose diameter is 10, 
is 17,320508; hence, 17,320508 
xX 10X 2=115,47005, the soli- 
dity required. 


Ex. 4. Required the solidity of an elliptical rectangular re- 
voloid, whose vertical axis is 48 inches, and conjugate axis is 
36 inches. 

36 X 36 X 48 X 2 =41472 cubic inches. 

Ex. 5. What is the solidity of an elliptical rectangular re- 
voloid, whose vertical axis is 36 inches, and whose conjugate 
is 48 inches ? Ans. 55296 cubic inches. 

Ex. 6. What is the solidity of an oblate spheroid, whose 
revolving axis = 48, and whose conjugate or fixed axis = 36 
inches ! Ans. 43429.4784 cubic inches. 


MENSURATION, ETC. 213 


Ex. 7. Required to find the solid content of the earth, sup- 
posing its circumference to be 25000 miles. 
Ans. 263859375000 cubic miles. 
Ex. 8. What is the solid content of a sphere, whose di- 
ameter AB = 25 feet? 
Ans. 81825 cubic feet. 
Ex. 9. Required the solidity of a sphere,4whose circumfer- 
ence is 18.6 {fcet. Ans. 108.665413272 cubic feet. 


PROBLEM. Il. 
To find the surface of a sphere or of a spherical revolovd. 


Rore.— Multiply the perimeter of its central conjugate sec- 
tion by the vertical axis, and the product is the whole surface. 

Ex. 1. What is the surface of a sphere whose diameter is 
10? Ans. 31.4159 x 10=814.159, the surface required. 


Ex. 2. Required the surface 
of a rectangular spherical revo- 
loid, BCED, whose vertical axis 
is 10 feet. 

-10X4x10=400 square feet, 
the surface required. 


Ex. 3. Required the surface of a ball, whose diameter AB 


= 1 inch. Ans. 3.1416 square inches. 
Tux. 4. How many square inches will cover a globe of 12 
inches in diameter ? Ans. 452.3904 square inches. 


Ex. 5. Required the superficies of the terraqueous globe, 
supposing the diameter AB = 7958 miles. And if only one- 
fourth part of its surface be dry land, and two acres sufficient 
to produce food for one person; how many persons can live 
on the earth at one time. 

198956786.5824 sq. miles, the surface of the globe 
Ans. 49739196.6456 sq. miles, dry land. 
15916542927 persons can live on the earth. 


7 ° 


214 MENSURATION 
PROBLEM III. 
To find the solidity of any segment or zone of a sphere. 


Rutz.—To half the sum of the areas of the two bases mul- 
tiplied by the altitude, add the solidity of a sphere whose dia- 
meter is equal to the altitude of the segment or zone. 

Ex. 1. What is the solidity of a_sphe- 
rical segment ABD, whose base is 10 and 
whose height oD is 21 


10X2=20. 

and 1¢7D*=1 x 3.14159X2x2x2=4,18876. | } 

hence 20+-4.18876=24,18876 = the \ vy, 
solidity of the segment. ele i Set 


Ex. 2. What is the solid content of a 
zone EFDCE, whose height or = 30 in- 
ches, the greater diameter EF = 60 in- , : 
ches, and pats diameter AB = 40 in- &f bo 7 
ches ? ns. 75398.4 cubic inches. ll, -—L WJ 

Ex. 4. Required the solidity of the mid- ee 
dle zone of a sphere ABDCA, the diam- 
eter of the whole sphere EF = 80 inches, 
the height nr=64 inches. : 

Ans. 233070.5408 cubic inches. 


\\\' 


PROBLEM IV. 


To find the convex surface of any segment or xone of a 
sphere, or spherical revoloid. 


Rute.— Multiply the perimeter of a middle section of the 
whole sphere or revoloid, perpendicular to the vertical axis, 
by the height of the segment or zone. 

Scholium. This is the same as the rule given in the Ele 
ments of Geometry for a spherical segment or zone, viz: its 
convex surface is there said to be equal to the height of the 
segment or zone, multiplied by the circumference of the 
sphere. The same rules as there given for segments and sec- 
tors of a sphere, will answer also for segments and sectors of 
right revoloids. 

Ex. 1. What is the convex surface of a segment of a right 
revoloid, whose height is 2 feet, the perimeter of a central 
conjugate section of the whole revoloid being 40 feet ? 

40 x2=:80, the surface required. 


OF SOLIDS. 215 


Ex. 2. Required the convex surface of a zone of a rectan- 
gular right revoloid whose height is 6 feet, the whole altitude 
of the revoloid being 10} feet. Ans. 252 feet. 

Ex. 3. Required the convex surface of a segment of a hex- 
agonal right revoloid, whose height is 54 feet, the axis of the 
revoloid being 10 feet. Ans. 190.5255825 square feet. 


PROBLEM V. 
Lo find the solidity of a sector of a spherical or right revoloid. 


Rure.—Multiply its convex surface by one-third the semi- 
axis of the revoloid. 


Kix. What is the solidity of a revoloidal sector, whose con- 
vex surface is 10 square feet, the axis of the revoloid being 
10 feet 1 

10x1xX5=162, the solidity. 


PROBLEM VI. 


To find the selidity of a segment or zone of a spherical revo- 
loid. 


Rute.—F ind the solidity of the revoloidal sector having the 
same convex surface ; find also, the solidity of the pyramid 
having the same base as the segment, and whose vertice is in 
the centre of the revoloid: subtract the solidity of the pyra- 
mid from that of the sector, which will give the solidity of the 
segment, if the segment is less than a semi-revoloid ; and add 
the solidity of the pyramid to that of the sector, if the seg- 
ment be greater than a semi-revoloid. 


Ex. 1. What is the solidity of a segment of a rectangular 
revoloid whose convex surface is 40 square feet, the axis of 
the revoloid being 10 feet ? 

Here, the sector will be found = 40x}=662 solid feet, 
and the height of the segment will be found = 1 foot; hence, 
(5’—4") =36 = the base of the segment; and 386X4X j= 48 
= the solidity of the pyramid. 

Therefore, 662--48=182 cubic feet the solidity required. 

Ex. 2. Required the solidity of the segment of an octagon- 
al revoloid, whose convex surface is 100 feet, the axis of the 
revoloid being 10 feet. 


216 MENSURATION 


PROBLEM VII. 


To find the solidity of a segment of a spheroid, made by a plane 
parallel to either azis. 


Scholium. Since the segment of a spheroid is the segment 
of a sphere, expanded or contracted in the ratio of the major 
and minor axes ; hence, we have the following. 


Rue. Find the solidity of a corresponding segment of the 
same altitude, from a sphere described on the same axis as 
that of the segment ; then, as this axis is to its conjugate, so is 
the spherical segment to the spheroidal segment, if the seg- 
ments base is parallel to the fixed axis. Or as the square of 
this axis is to the square of its conjugate, so is the spherical 
segment to the spheroidal segment, if the base is circular of 
parallel to the revolving axis. 


Nore. The same will also apply to the segment of an ellip- 
tical revoloid compared with a corresponding segment of a 
spherical or right revolvid. 


Ex. 1. In the prolate spheroid 
ACBD the fixed axis AB=50. the re- 
volving axis CD=80, required the so- 
lidity. of the segment EFCE, its 
height EG=6, the base being parallel 
to the fixed axis AB., 


The solidity of a spherical segment, whose height is CG, 
the diameter being CD=1470.2688. 

Hence, by the rule, 

CD: AB: : 1470.2688 : EFCE; 
or 30 : 50: : 1470,.2688 : 2450.448 the solidity of the segment 
EFCE. 


Ex. 2. In an oblate spheroid, whose revolving axis AB=50, 
the fixed axis CD=30; required the solid content of the seg- 
ment EFCE, whose height = 5, its base being perpendicular 
to the revolving axis. Ans. 1099.56. 


Ex, 3. Required the solidity of the segment EFCE of the 
prolate spheroid ACBDA, the fixed axis DB=48, the revolving 
axis CD=88. and the height of the segment CG=16, the base 
being perpendicular to the revolving axis. 

Ans. 13883.8878. 


UN eee 


OF SOLIDS. 217 


Hx. 4. Required the solidity of 
the segment EAFE of a prolate 
spheroid, the height AG=5 inches, 
its base being parallel to the revolv- 
ing 8 which is 30 inches, its fixed 
_ axis being 50 inches. 


BD 
The solidity of a spherical segment, whose altitude is AG 
of a sphere, whose axis is AB, is 1832.6 cubic inches. 
AB? : CD? : ; 1882.6: EAFE ; 
or 2500 : 900 ; : 1832.6 : 659.736 the solidity of the segment 
EAFE. | 


Ez. 5. Required the solid content ofthe segment of the prolate 
spheroid EAFE, its base being parallel to the revolving axis ; 
the height AG=1, the fixed axis AB—10, and the revolving 
axis CD=6., Ans. 5.2778’, 


fiz. 6. The fixed axis CD of an obiate spheroid being 30, 
the revolving axis AB=50, and the height of the segment=6, 
its base being parallel to the revolving axis ; required the so- 
lidity of the spheroidal segment. Ans. 4084.08. 


PROBLEM VIII. 
To find the solid content of the middle frustum of a spheroid. 
CASE I. 
When the ends are circular, or parallel to the revolving azis. 


Rute. To twice the square of the middle diameter, add the 
square of the diameter of oneend; multiply this sum by the 
length of the frustum, and the product again by .2618 (which 
as one-third of .7854,) for the solidity of the middle frustum. 


Scholium. This, and the following rule is derived from the 
principles contained in scholium page 156, this volume. 


Ex. 1. Required the solidity of 
the middle frustum FCGHDFE ofa 
prolate spheroid, the middle diame- 
ter CD=30, the diameter of each 4 
circular end EF or GH=18, and 
the length cr=40, 


D 
Here (2CD*+GH?’) x cr X .2618 = [(30?X 2)+ 18°] X 40X 
.2618=(18004-324) X 40X .2618 = 22242.528 the solidity of 
the middle frustum ECGHDFE required. 
15 


é 


218 MENSURATION 


Ex. 2. What is the solidity of the middle frustum 
ECGHDFE of an oblate spheroid, having the less diameters 
of the circular ends EF and GH, each equal 40; the middle 
or greater diameter CD=50, and the length cr=18 ? 

Ans. 31101.84. 


CASE Il. 


When the ends are elliptical, or perpendicular to the revolving 
axis. 


Ruz. To twice the product of the major and minor axes 
of the middle section, add the product of the major and minor 
axis of one end ; then multiply this sum by the length of the 
frustum, and the product again by .2618, for the solid content 
of the middle frustum. 


Haz: 4. Inthe middle frustum 4% 977-7: 2 


. 0» Sar teed ht 
EFHGE of an oblate spheroid, the i = > >», 
major and minor axes of the middle Al& | = A 


or greater elliptic section AB are 50 yy 
att, 30, and ay major and minor “Es a 4 
axis'at one’ end EF are 40 and;24,° @ “**-~_4. 
the height IK=9 ; required the solid content of we middle 
frustum. 

Here (50X30X2) + (40X24) x 9 X .2618= (300+960) x 
2.3562 = 9330.552, the solidity of the frustum EFHGE re- 
quired. 


Ex. 2. In the middle frustum EFHGE of an oblate spheroid, 
the two axes of the middle ellipse are 50 and 80, and those of 
each end are 30 and 18, the height of the frustum IK = 40 ; 
required the solid content of the frustum KFHGE. 

Ans. 37070.88. 


PROBLEM IX. 


To find the solidity of a paraboloid or a vertical parabolic 


revoloid. 


Rute. Multiply the area of the base by half the height. 


Ex. 1. Ifthe diameter of the base of 
a paraboloid be 12 feet, and height 22 
feet, what is the solidity ? 
Ans. 1243. 


OF SOLIDS. 219 
Ex. 2. If the sides AB, CB of the rectangu- 


lar base of a parabolic semi-revoloid or pyra- 
moid ABCD are each = 10 inches, and the al- 
titude I-D=18 inches, required its solidity. ; : 

Ans. 1800 cubic inches. =a 


PROBLEM X. 


To find the solidity of the frustum of a parabolic conoid, or 


or paraboloid, or of a vertical parabolic revoloid. 


Roce. Multiply the sum of the areas of the two ends by 
half their distance. 


Ez. 1. What is the solid content of 
the frustum of a paraboloid, the greater 
diameter DC = 30, the least diameter 
AB=24, and the altitude EF=9 ? 

Ans. 5216.6268. 


Ex. 2. What is the content in wine gallons of a cask in the 
form of two equal frustums of a paraboloid ; the length=2EF 
=40 inches, the bung diameter DC=32 inches, and the head 
diameter AB=24 inches ; the gallon containing 231 cubic 
inches ? Ans. 108.768 gals. 


PROBLEM XI. 


To find the solidity of a hyperbolic conoid, or otherwise called 
a hyperboloid. 


Rute. To the square of the radius of the base, add the square 
of the diameter in the middle, between the base and top; multi- 
ply this sum by the altitude, and the product again by .5236, 
for the solidity of the hyperboloid. (Art. 22, Chap. I, B. V.) 


iz. 1. What is the solidity of an hyperboloid MBNM, 
whose altitude KB=10, the radius of its base MK=12, and 
the middle diameter PG=6 7? 


2230 MENSURA TION: 


Here (MK? + PG’) x KB x .5236 
= [122 + (6v7)] X 10 xX 5.236 
= [144 + ( X 7)] X .5236 = 
(144 + 252) x 5.236 = 2073.456, the 
solidity of the hyperboloid MBNM re- 
quired. 


RE og 


Ex. 2. Required the solidity of the hyperboloid MBNM, 
whose altitude KB=50, the radius: of its base MK=52, and 
the middle diameter PG=68. Ans. 19184'7.04., 


PROBLEM, XII. 


To find the solidity of the frustum of a hyperbolic conoid, or 
hyperboloid. 


Rots. To four times the square of the middle diameter, add 
the sum of the squares of the greatest and least diameters ¢ 
then multiply this sum by the altitude of the frustum, and that 
product again by .1309, (et being the sixth part of .7854,) for 
the solidity.. 


Ex. 1. Required the solidity of the & 
frustum of the hyperbolic conoid 


PGDCP, the height EH=7, the great- 


est diameter CD=48, the middle diam- 
eter MN=38, and the least diameter Bi 
PG=27. 

Here (4MN?+CD? + PG’) x EHx mn 
.1309=[ (38° 4)+48°+277] X .1309= 4, A? _\\\\\w 


(5776+ 2304+-729) X7X .1309 = 8809 
x 7X.1309==8071.6867, the solid com- cg 
tent of the frustum PGDCP required. 


Lz. 2. Required the solidity of the frustum ‘of a hyperbo- 
lic conoid PGDCP, whose greatest diameter CD = 10, the 
least diameter PG=6, the diameter MN=8}, and the altitude 
BEH=12. Ans. 667.59. 

fiz, 3. A cask, in the form of two equal frustums of a hy- 
perbolic conoid, having its bung diameter CD=32 inches, its 
head diameter PG = 24 inches, and the diameter in the mid- 
dle, between the bung and head MN=£,/310, the length of 
the cask 2 EH = 40 inches; required the content in wine 
gallons. 

Ans. 24998.7584 cubic inches= 108.219, &c. gallons. 


OF SOLIDS. 224 
PROBLEM XII. 


Po find, by a general rule, the solidity of any-solid, frustum, or 
segment, produced by the revolution of any conic section, or 
of any revoloid circumseribing such solids. 


GeneraL Rute. To the sum of the ends, ‘add four times 
‘a section equidistant therefrom, and multiply this sum by one- 
‘sixth of the length. 


Scholium. This rule is true for any solid or segment, which 
is generated by any multiple or power of a-series of numbers 
in arithmetical progression. (See Beok Y, Chap. 1 & 2.) 


Ezx.1. What is the solidity of a-sphere, whose diameter is 2? 

The area ef its central section, or of its great circle is 
3,14159. 

Hence 3,14159x 4x 2=4,1887%8 the solidity. 

ix, 2. What isthe solidity of a zone of a spheroid, whose 
two bases are10 and 5 square inches, and whose central section 
parallel to the bases is 9 square inches, the height ef the zone 
being 18 inches ? Ans. 1538 cubic inches. 


‘PROBLEM XIV. 


To find the solidity of & circular spindle, produced by the revo- 
lution of a@ circular segment about its base er chord as an 
avis. 


Rots. From tof the cube ofhalf the axis, subtract the pro- 
duct of the central distance into half the revolving circular 
segment, and multiply the remainder by four times 3.14159. 

If a=the area of the revolving circular segment, 

/=half the length-or axis of the spindle, 
[c=the distance of the axis from the centre of the circle 
. to which the revolving segment belongs ; 
The solidity = (1’—1ae)X 4X 3.14159. Cc 


Ez, Wet a circular spindle ACBD be pro- 
duced by the revolution of the segment 
ABC, about AB. If the axis AB be 140, 
and CP half the middle diameter of the 
spindle be 38.4 ; what is the solidity ? 


The area of the revolving segment is 3791 
The central distance OF 44,6 
The solidity of the spindle 374402 . 


222 MENSURATION 
PROBLEM XV. 
To find the surface of a circular spindle. 


Rute. Subtract the product of the revolving arc, multiphed 
by the central distance, from the product of the length of the 
spindle into the radius, and multiply the remainder by twice 
3.1416, and this product will be the surface. 

Ex. Required the surface of a circular spindle ACBDA, 
whose length AB = 40, and middle diameter CD. = 30, the 
length of the arc ACB being 531. 

The height of the revolving segment=15, the radius of the 
circle==202, the central distance OM =55. 

(AB x OC—ACB x OE) X 3,1416 2 
= (40 X 203—531 X 5) 6.2832= 3281. ta ll the surface of. the 
spindle ACBDA required. 


PROBLEM XVI. 
To find the solidity of the middle frustum of a circular spindle. 


Rute. From the square of half the axis of the whole spin- 
dle subtract $ of the square of half the length of the frustum ; 
multiply the remainder by this:half length ; from the product 
subtract the product of the revolving area into the central dis- 
tance ; and multiply the remainder by twice 3.14159. 

If L=half the length or axis of the whole spindie, 

J=half the length of the middle frustum, 
c=the distance of the axis from the centre of the circle, 
a=the area of the figure which, by revoling, produces 
the frustum ; 
The solidity =((L’—}?) x ac) X2X3.14159. 


Ex. Hf the diameter of each 
end of a frustum of a circuiar 
spindle be 21.6, the middle diam- 
eter 60, and the length 70; what 
is the solidity ? 


C 
The length of the whole spindle is 79.75 
The central distance 11.5 
The revoling area 1703.8 
The solidity 136751.5 


Scholium. The middle frustum of a circular spindle may be 
resolved into a cylinder, whose two bases are AD, CB, and a 


OF SOLIDS: 223 


ring described by the revolution of the segment CDE about the 
axis KL; this ring may also be resolved into a cylindric seg- 
ment, whose base is the segment DCE, and whose altitude is 
=the inner diameter of the ring, and a circular spindle formed 
by revolving the segment DCE about its chord DC. (Prop. 
XJ, Cor. 3, B. Il.) Hence its content may be calculated ac- 
cordingly. 


PROBLEM XVII. 
To find the solidity of an elliptic spindle. 


Rute. First find the solidity of a circular spindle, generated 
from a segment, whose height CG, is the same as that of the 
elliptical. segment,.generating the elliptic spindle, the radius of 
the circle being CO. 

Then, as the length of the circular spindle is to that of the 
elliptic spindle, so is the solidity of the circular spindle to that 
of the elliptic spindle. 


Fiz. If half the middle diameter 
CD of anelliptic spindle is 38.4, and 
its axis AB=200, its central distance 
OG being 44.6, what isits solidity ? 1 

The solidity of a circular spindle 
having the same middle diameter, | 
and the same central distance, we H- 
have found (Prob. XIV,)=874402, but its length is 140, there- 
fore by the rule. 

140 : 200 : 374402 : 534860 the solidity required. 


PROBLEM XVIII. 


To. find the solidity of a parabolic spindle, produced by the re- 
volution of a parabola about a double ordinate or base. 


Roe. Multiply the square of the middle diameter by ;4; of 
the axis, and the product by .7854., 


Ex. If the axis of a parabolic 
spindle be 30, and the middle 
diameter 17, what is the solidity ? 

Ans. 3631.7 


PROBLEM XIX. 
To find the solidity of the middle frustum of a parabolic spindle. 


Ruts. Add together the square of the end diameter, and 
twice the square of the middle diameter ; from the sum sub- 


224 MENSURATION 


tract 2 of the square of the difference of the diameters, and 
multiply the remainder by + of the length, and the product 
by .7854. 

If D and d=the two diameters, and /=the length ; 

The solidity = (2D’+d?—2 (D—d)’) x j/X.7854. 


Ex. If the end diameters of a 
frustum ofa parabolic spindle be 
each 12 inches, the middle diam- 
eter 16, and the length 30; what 
is the solidity ? 

Ans. 5102 inches. 


PROBLEM XX. 
To find the convex surface of a cylindric ungula. 


Ruz. From the product of the diameter and sine, subtract 
the product of the arc and cosine, and multiply the difference 
by the altitude divided by the versed sine. 

Let h=the altitude AD, 
v=the versed sine AF, 
d=the diameter AB 
a=the arc EAG, 
s=the right sine FG, 
c=the cosine of the half are. 


ds—ac 
Then—-— x h=the convex surface. 


Scholium 1. When F is the centre of the base ; then v=s= 

d,c=o ; and then the rule becomes dh, viz., the surface is= 
the product of the diameter into the height. 

2. When AF excedes + AB, then ac must be added, and the 


ds-+-ac 
expression becomes — xh=the surface. 


Fix. 1. What is the curve surface of an ungula EGDA, 
whose base is half the base of the cylinder and height, AD= 
10, the radius FG=10? Ans. 100. 

Hz. 2. Given the diameter AB = 100 the height AD=140, 
and the versed sine AF=10, required the curve surface. 

Ans. 5962,738. 


Scholium. The same considerations will apply to cylindric 
ungulas, as for circular spindles, taking GE as the axis of 
the spindle, and AD the circumference of a middle section. 
(Prop. IV, Cor. 1, B. III.) 


OF SOLIDS. 225 
. XXL, 
To find the solidity of a cylindric ungula. 
CASE I. 
When the base of the ungula is = half the base of the cylinder 
Ruts. Multiply the square of the radius of the base by the 


altitude of the ungula, and take 2 the product for the solidity. 


Scholium. This rule is absolute, without 
reference to the circle’s quadrature, (Prop. 
VI, Cor. 5, B. IIL) (s=27°h, Formula 6, Page 
92.) 


iz. What is the solidity of the cylindric 
ungula AE FC, whose base EFC is half that 
of the cylinder, the diameter EF being 6, and 
the altitude CA of the ungula being 16, 
3°X 16X2=96 the solidity required. 


CASE II. 


When the base of the ungula is greater or less than half 
that of the cylinder. 


Subtract the product of the area of the base by the differ- 
ence between the radius and the versed sine or height, from 
one-twelfth of the cube of the chord of the base, if the versed 
sine be less less than the radius, otherwise add this product, 
multiplying this result by the altitude ot the ungula, and 
divide this product by the versed sine. 

If a=the area FEC of the segment forming the base of the 
ungula, r=the radius, FI v=the versed sine CI,c = the chord 
EP, and A=the altitude AC. 

Then will the solidity of any cylindric ungula= 


(ge ur N v) a) i. 


Ex. Given the diameter HC 50 inches, the altitude AC of 
the ungula=120 inches the versed sine IF =10 inches, requir- 


ed the solidity of the ungula. 
The chord EF will be found=40 inches. The area of the 


base 279.56. 
Whence, by the rule, AF being less than } HC, we have 


(40° (rv) a) =136709} cubic inches, the solidity of the 
ungula EFCA. 


226 MENSURATION 


Scholium. When the section passes ob- 
liquely through the opposite sides of the 
cylinder, the content of the ungula may be 
found by multiplying the sum of the greatest 
and least heights of the ungula by the area 
of the base, and its surface may be found by 
multiplying ithe sum of the greatest and least 
heights by the perimeter of the base. 


Hence, the ungula ABD is equal to half pg 
the cylinder ABCD, both in its surface and | 
solidity. 


2. The complement LMHKD, DKHCB | 
of any ungula LMPD, BGAED, may be ? 
found by subtracting the solidity of the un- 
gula from a portion of the cylinder of equal 
base and altitude. 


| oy il Nh 
IA 
PROBLEM XXII. 


To find the solidity of a conical ungula. cut from the cone, or 
frustum, by a plane parallel to-the side of the cone. 


Rute. Multiply the area of the base by the diameter of the. 
base of the frustum, and divide the product by the difference 
of the diameters.of the two bases ; fom this quotient subtract 
four-thirds of the product of the less diameter by the square 
root of the product.of the less diameter, and difference of the 
diameters. Multiply the remainder by one-third of the height, 
and the product will be the content. 

Let a=the area.of.the base cmB, 

D=AB the diameter of the base of the 
frustum, 

d=KD the diameter at the top, 

h=the height. 

Then will the solidity of the ungula= 


aD 
(Fy-tex vd(D— a) th 


Ex. If the diameter AB=30 inches, the diameter ED=19.2 
inches, and the height od = 18 inches, what is the content of 
the ungula ? Ans. 1606.41. 


OF SOLIDS. 227 


Scholium. Other formule may be found in the general 
scholia, at pages 191, &c., for cylindric and conie ungulas; it 
is unnecessary to extend the problems here. 


PROBLEM XXIII. 
To find the solidity of a cylindrical ring. 


Rute.. Multiply half the sum of the inner and outer circum- 
ferences by the area of a section of the ring. 


Scholium. This rule answers for all rings, the virtual centres 
of whose sections are in the centre of magnitudes of such 
sections. 


For all other rings see Prop. XI. and Corollaries B. IIL. 


What is the solidity of the cylindric 
ring AD, whose inner diameter BC | 
is 10 inches. and whose outer diam- 4 
eter AD is 20 inches ? 

Ans. 925.436 inches. 


Scholium. This ring is equivalent to a segment: ABGFA 
ef a cylinder, whose section DE is equal to that of the ring, 
and whose length FG = the inner circumference, and AB = 
the outer circumference. 


——S—SS= 


SaaS 


K i) ot ee Be 


Ex. 2. What is the solidity of the cyl- 
indric ring EF, whose inner diameter is 
0, and its outer diamer EF 10 inches? 

Ans. 308,478 inches. 


SS g 


Scholium. This ring is equivalent to the segment ABC of a 
cylinder, whose length AB is=the outer circumference of the 
ring. 


=o, 
= 
= 
SI 
= 


f 
41 
Lat 


GAUGING OF CASKS. 


Art. 1. Gauging of casks is a practical art; and since 
casks are not commonly constructed in exact conformity with 
any regular mathematical figure, the subject does not admit 
of being treated in a very scientific manner ; by most writers 
on the subject, however, they are considered as nearly coin- 
ciding with one of the following forms : 


het) ig of a spheroid, 
2: | Llsiisetra tenes adlee, of a parabolic spindle. 


of a paraboloid, 


The equal frustums 
of a cone. 


4. 


__The second of these varieties agrees more nearly than any 
of the others, with the forms of casks, as they are commonly 
made. The first is too much curved, the third too little, and 
the fourth not at all, from the head to the bung. 

2. Rules have already been given, for finding the capacity 
of each of the four varieties of casks. As the dimensions are 
taken in inches, these rules will give the contents in cubic 
inches. ‘To abridge the computation, and adapt it to the par- 
ticular measures used in gauging, the factor .7854 is divided 
by 282 or 321; and the quotient is used instead of .7853, for 
finding the capacity in ale gallons or wine gallons. 


, 7854 
Now Ja5 = 002785, or .0028 nearly ; 


7854 
And 93] 70084 


If then .0028 and 0084 be substituted for .7854, in the rules 
referred to above ; the contents of the cask will be given in 
ale gallons and wine gallons. These numbers are to each 
other nearly as 9 to 11. 


PROBLEM I. 


To calculate the contents of acask, in the form of the middle 
frustum of a spheroid, being a cask of the first variety. 


Rute. Add together the square of the head diameter, and 
twice the square of the bung diameter; multiply the sum by 
1 of the length, and the product by .0028 for ale gallons, or 
by .0034 for wine gallons. 


GAUGING. 229 


If D and d= the two diameter, 
and /=the length ; 


The capacity in inches= 
(2D? +-d?) X 1X .7854. 


And by substituting .0028 or 
.0034 for 7854, we have the capacity 
in ale gallons or wine gallons. 


Ex, What is the capacity of a cask of the first form, whose 
Jength AB is 80 inches, its head diameter EF 18, and its 
bung diameter CD 24 ? | 
Ans. 41.3 ale gallons, 

or 50.2 wine gallons, 


PROBLEM Ii. 


To calculate the contents of a cask, in the form of the middle 
frustum of a parabolic spindle, being a cask of the second 
variety. 


Rute. Add together the square of the head diameter, and 
twice the square of the bung diameter, and from the sum sub- 
tract 2 of the square of the difference of the diameters ; mul- 
tiply the remainder by 1 of the length, and product by .0028 
for ale gallons, or .0034 for wine gallons. 


The capacity in inches= 


C 
(2D?-+-d'—2 (D—d)") x 11x.7854. ZZ | 


Ex, What is the capacity of a 4) 
cask of the second form, whose i 
length AB is 30 nches, its head diam- . 
eter BF = 18, and its bung diameter 
CD=24 7 


Answer 40.9 ale gallons, 
or 49.7 wine gallons. 


PROBLEM IiIl. 


To calculate the contents of a cask, in the form of two equal frus- 
tums of a paraboloid, being a cask of the third variety. 


Ruxe. Add together the square of the head diameter, and 
the square of the bung diameter ; multiply the sum by half the 
length, and the product by .0028 for ale gallons, or .0084 for 
wine gallons. 


230 GAUGING. 


The capacity in inches= 
(D?+d?) X11x.7854. 


Ex. What is the capacity of a cask 
of the third form, whose dimensions 
are,as before, 30, 18, and 24? “D 
Ans. 37.8 ale gallons, 
or 45.9 wine gallons. 


SS gs 


PROBLEM IV. 


To calculate the contents of a cask, in the form of two equal 
Srustums of a cone. | 


‘Rute. Add together the square of the head diameter, the 
square of the bung diameter ; and the product of the two diam- 
eters ; multiply the sum by 3 of the length, and the product by 
0028 for ale gallons, or .0034 for wine gallons. 

Tne capacity in inches= 
(D?+d?+ Dd) x 11X.7854. 


Fix. What is the capacity of a cask 4{)) 
of the fourth form, whose length AB is Aji 
30, and its diameters EF and CD = 18 \b 


and 24? 


Ans. 37.3 ale gallons, 
or 45,3 wine gallons. 


Scholium. In the preceding rules, it is supposed that the cask 
corresponds to the different varieties, whereas, it is seldom that 
a cask perfectly coincides with either ; but for the greater 
certainty of the truth, when accuracy is required, the follow- 
ing rules, the demonstration of which will be found in Hut- 
ton’s Mensuration, are to be preferred. 


PROBLEM V. 


To calculate the contents of any common cask from three 
dimensions. 


Ruue. Add together 
25 times the square of the head diameter, 
39 times the square of the bung diameter, and 
26 times the product of the two diameters. 
Multiply the sum by the length, divide the product by 90, 
and multiply the quotient by .0028 for ale gallons, or .0034 for 
wine gallons. 


GAUGING. 231 
The capacity in inches= 
l 
(39D +25d" + 26Dd) x — x.7854. 
Ex. What is the capacity of a 
cask whose length is 30 inches, the 


head diameter 18, and the bung 
diameter 247 


Ans. 39 ale gallons, 
or 471 wine gallons. 


PROBLEM VI. 


To calculate the contents of a cask from four dimensions, the 
length, the head and bung diameters, and a diameter taken 
in the middle between the head and the bung. 


Rus. Add together the squares of the head diameter, of 
the bung diameter, and of double the middle diameter ; multi- 
ply the sum by } of the length, and the product by .0028 for 
ale gallons, or .0034 for wine gallons. 

If D = the bung diameter, d = the head diameter, m = the 
middle diameter, and /=the length ; 

The capacity in inches= 


(D?+d?+2m’) X4/X.7854. 
A 1 
Ex. What is the capacity of a qj 
cask, whose length cr is 30 in- ™ 
ches, the head diameter EF=18, 
the bung diameter CD = 24, and the middle diameter IK 22}? 


Ans. 41 ale gallons, 
or 492 wine gallons. 


Scholium. In making the calculations in gauging, accord- 
ing to the preceding rules, multiplications and divisions are 
frequently performed by means of a Sliding Rule, on which 
are placed a number of logarithmic lines, similar to those on 
Gunter’s Scale. 

Another instrument commonly used in gauging is the Diag- 
onal Rod. By this, the capacity of a cask is very expeditiously 
found, from a single dimension, the distance from the bung 
to the intersectiou of the opposite stave with the head. The 
measure is taken by extending the rod through the cask, from 
the bung to the most distant part of the head. The number 
of gallons corresponding to the length of the line thus found, 
is marked on the rod. The logarithmic lines on the gauging 
ae are to be usad in the same manner, as on the sliding 
rule. 


232 GAUGING. 


ULLAGE OF CASKS. 


Art. 2. When a cask is partly filled, the whole capacity is 
divided, by the surface of the liquor, into two portions ; the 
least of which, whether full or empty, is called the udlage. 
In finding the ullage, the cask is supposed to be in one of two 
positions; either standing, with its axis perpendicular to the 
horizon ; or dying, with its axis parallel to the horizon. The 
rules for ullage which are exact, particularly those for lying 
casks, are too complicated for common use. The following 
are sufficiently near approximations. See Hutton’s Mensura- 
tion. 

PROBLEM VII. 


To calculate the ullage of a standing cash. 


Ruts. Add together the squares of the diameter at the sur- 
face of the liquor, of the diameter of the nearest end, and of 
double the diameter in the middle between the other two; 
multiply the sum by } of the distance between the surface and 
the nearest end, and the product by .0028 for ale gallons, or 
.0034 for wine gallons. 

If D=the diameter of the surface of the liquor, 

d=the diameter of the nearest end, 
m=the middle diameter, and 
l=the distance between the surface and the nearest end; 


The ullage in inches= (D*+d?+-2m’) X3/1X.7854. 


Ex. If the diameter at the surface of the liquor, in a stand- 
ing cask, be 32 inches, the diameter of the nearest end 24, the 
middle diameter 29, and the distance between the surface of 
the liquor and the nearest end 12; what is the ullage ? 

| Ans. 274 ale gallons, or 33? wine gallons. 


PROBLEM VIII. 
To calculate the ullage of u lying cask. 


Rue. Divide the distance from the bung to the surface of 
the liquor, by the whole bung diameter, find the area of a cir- 
cular segment, whose versed sine is the quotient in a circle, 
whose diameter is 1, and multiply it by the whole capacity of 
the cask, and the product by 14 for the part which is empty. 

If the cask be not half full, divide the depth of the liquor by 
the whole bung diameter, and find the area of the segment, 
multiply, &c., for the contents of the part which is full. 

Ez. If the whole capacity of a lying cask be 41 ale gallons, 
or 492 wine gallons, the bung diameter 24 inches, and the dis- 
tance from the bung to the surface of the liquor 6 inches, what 
is the ullage ? Ans. 7? ale gallons, or 9} wine gallons. 


OF THE 
SPECIFIC GRAVITY OF SOLIDS AND FLUIDS. 


Tue specific gravities of bodies are their relative weights, 
contained under the same given magnitude as a cubic foot, or 
a cubic inch, &e. 


A table of the Specific Gravities of Bodies, and the weight of 


a cubic foot of each, in ounces, avoirdupois. 


Platinum, Common stone, . . 2520 
Rolled, . ‘ 22666 Loom, . ’ & 2160 
Hammered, . 20335 | Clay, .. : : 2160 

Pure gold Brick, . ; j 2000 
Hammered, . 19360 Ivory, °. . - 1825 


Cast, : 19256 Sand, . : : 1520 
Gold 22 car. fine cast, Coal, ; : ET 
‘17484 | Sulphuric acid, . 1840 
Mercury, ; 13596 | Nitrous acid, . 1550 
Lead; . A a Nitric acid, . : 12s 
Silver, cast, : 10474 Human blood, : 1054 
Copper, 4 . ° 8788 | Cow’smilk, . - 1081 
Soft steel, Box-wood, . F 1030 
Hammered, . 7839 Sea-water, . “ 1028 
MOaste. ©. ; 7832 Vinegar, . : 2 lO0eg 
Hard steel, ‘are’ wi ; : 1015 
Hammered, , 7817 Common water, . 1000 
Case ee : 7815 Red wine, . : 990 
Barwon... et T7987 1.’ Linseed oil, *. blah! ci 3e- 
Tin, : : 7290 Proof spirits at 510, 923 
Cast iron, | % : 7208 Olive oil, . ; 913 
Zinc, . , 6860 Alcohol, pure, . Toe 
Granite, . 3500 to 4000 fither, ~’. s TOG 
Flint glass, . ‘ 3329 Wir, ‘ : 12 


Note. The several sorts of wood are supposed to be dry. 
Also as a cubic foot of water weighs just 1000 ounces, avoir- 
dupois, the numbers in this table express not only the specific 
gravities of the several bodies, but also the weight of a cubic 
foot of each, in avoirdupois ounces ; and hence, by proportion, 
the weight of any other quantity, or the quantity of any other 
weight, may be known as in the following problems. 

16 


234 MAGNITUDES AND 


PROBLEM. I. 
To find the magnitude of any body from its weight 


Rute.—As the tabular specific gravity of the body is to its: 
weight in avoirdupois ounces; so is’ one cubic foot, or 1728 
cubic inches, to its content in feet, or inches respectively. 

Ex. 1. Required the solid content of an irregular block of 
common stone, which weighs 1 cwt. or 1792 ounces. 

Here, as 2520 oz: 1792 ox: : 1728 cubic inches: to its so« 
lid. content. 

Or, 

As 5 oz. : 256 0z:: 24 cubic inches : 12284 cubic inches, 
the solid content required. 

Ex. 2. How many cubic feet are there in a ton. weight of 
dry oak ? ' Ans. 38138 cubic feet. 

Ex. 38. What is the solid content, and diameter of a cast 
iron ball, that weighs 42 pounds, its specific gravity being 
7208 ? 

aes Solidity = 161.095 cubic inches, 
" / Diameter = 6.75 inches.. 


PROBLEM II. 
To find the weight of a body from its:magnitude. 


Rurs.—As one cubic foot, or 1728 cubic inches, is to th 
solid content of the body, so is its tabular specific gravity to» 
the weight of the body. 

Ex. 1. Required the weight of a block of marble, whose 
specific gravity being 2700, the length.= 63 feet, the breadth 
and thickness each = 12 feet; this block being the dimensions» 
of one of the stones in the walls of Balbeck. 

Here, as 1 cubic foot : 68 X 12 X 12 (= 9072 cubic feet) 
: : 2700 oz. : 683,74 tons. weight, almost equal to the burthen 
of an East India ship. 

Ex. 2. What is the weight of a block of dry oak, which, 
measures 10 feet long, 3 feet broad, and 2} feet deep? 

Ans. 43351 pounds. 

Ex. 8. What is the weight of a leaden ball, 41 inches in 
diameter, its specific gravity being 11351 ? 

Ans. 16 Ibs. 7 oz. 

Ex. 4. Required the weight of a cast iron shell, 3 inches 
thick, its external diameter being 16 inches, and its specific 
gravity 7208. 

‘A Solidity = 1621.0656 cubic inches. 
'( Weight = 2 cwt. 3 qr. 9 lb. .5 oz. 


SPECIFIC GRAVITY. | 235 
PROBLEM II. 
To find the specific gravity of a body heavier than water. 


Rute.—Weigh the body both in water and out of water, by 
a hydrostatic balance, and take the difference of these results, 
which will be the weight lost in water. 

Then say, as the weight lost in water, is to the weight of 
the body in air, so is the specific gravity of water, to the spe- 
cific gravity of the body. 

Ex. 1. A piece of stone weighed ten pounds in air; but, 
in water, only 62 pounds ; required the-specific gravity. 

Here, as 10 — 62 (= 31) : 10: :.1000: to the specific gra- 
vity of the body. 

Or 


As 13 lbs. : 40 Ibs. : ; 1000 oz. : 3077 oz. = Ans. 

Ex. 2. A piece of copper weighs 36 oz. in air, and only 
31.904 oz. in water ; required the specific gravity of copper. 
. Ans. 8788 ounces. 

Eq. 2. Required the specific gravity of a piece of granite 
stone which weighs 7 lbs. in air, and 5 Ibs. in water. 
Ans. 3500 ounces. 


ante f PROBLEM IV. 
To find the specific gravity of a body lighter than water. 


Rutz. —Fasten to the lighter body, by a slender thread, ano- 
ther body heavier than water, so that the mass compounded 
of the two may sink together. Weigh the heavier body, and 
the compound mass, separately, both in water and out of it, 
then find how much each loses in water, by subtracting its 
weight in water from its weight in air. 

Then say, as the difference of these remainders is to the 
weight of the lighter body iz air, so is the specific gravity of 
water to the specific gravity of the lighter body. 

xx. 1. Suppose a piece of elm weighs 12 lbs. in air, and 
that a piece of metal, which weighs 18 Ibs. 7m azr, and 16 Ibs. 
in water, is affixed to it, and that the compound weight is 6 Ibs. 
in water ; required the specific gravity of the elm. 

Here 18 — 16=2 pounds, the metal lost 7m water ; 
and (18+15) —6 = 33 —6=27 pounds, the compound lost 
in water. 

Then 27 —- 2=25 pounds, the elm lost in water. 

As 27 — 2 (=25 Ibs.) : 15 Ibs. :.: 1000 oz. : to the specific 
gravity of the elm. 


x “sith 


236 MAGNITUDES, &c. 


| On, 

As 1 lb. : 3 lbs. : : 200 oz. : 600 oz. = Ans. 

Ex. 1. A piece of ash weighs 20 lbs. in air, to which is 
affixed a piece of metal, which weighs 15 lbs. in air, and in 
water, 131 lbs. ;,and the compound, in water, weighs only 84 
Ibs ; required the specific gravity of the ash. 

Ans. 800 ounces. 

Ex. 2. Suppose a piece of fir weighs 11 Ibs. in air, and a 
piece of steel being affixed which weighed 16 lbs. in air, and 
an water 14 lbs.; and the compound iz water weighs only 5 

lbs. ; what is the specific gravity of the fir? 
Ans. 550 ounces. 

Ex. 4. A piece of cork weighing 20 lbs. in air, had a piece 
of granite fixed to it, that weighed 120 lbs. in air, and 80 lbs. 
an water; the compound mass weighed 162 Ibs. in water ; 
what was the specific gravity of the cork ? 

Ans. 240 ounces. 


QUESTIONS FOR EXERCISE. 


1. Having a rectangular marble slab, 58 inches by 27, I 
would have a square foot cut off parallel to the shorter edge ; 
I would then have the like quantity divided from the remain- 
der, parallel to the longer side ; and this alternately repeated, 
till there shall not be the quantity of a foot left; what will be 
the dimensions of the remaining piece ? 

Ans. 20.7 inches by 6.086. 

2. Given two sides of an obtuse angled triangle, which are 
20 and 40 poles; required the third side, that the triangle may 
contain just an acre of land? Ans. 58.876 or 23.099. 

3. The ellipse in Grosvenor-square measures 840 links 
across the longest way, and 612 the shortest, within the rails ; 
now the walls being 14 inches thick, what ground do they in- 
close, and what do they stand upon ? 

A inclose 4a. Or. 6p. 
stand on 17601 sq. feet. 

4, What is the length of a chord, which cuts off one-third 
of the area, from a circle whose diameter is 289 ? 

. Ans. 278.6716. 


QUESTIONS FOR EXERCISE. 237 


A 


A st a ¥en a 


f 
! 
\ 
\ 


5. What is the area of the heart 
CABDFEC, the axis CD = 10 inches. 
(See article on spirals, page 140.) \ ‘ 

Ans. 104.7198 inches. SE 


oo 


6. There are two pul- 
leys AHD, BFE, the di- 
ameters AD, and BE 
are each = 20 inches, 
and the distance Cc is 
4 feet; the pulley BFE 
is put in motion around 
its axis by a belt ABFE-* 
DH, passing round AHD. Now if the pulley ILK on the 
same axis with AHD, is 40 inches in diameter, what must te 
the diameter hi of a corresponding pulley hft, around which 
the belt I/feK L may pass, so as to be of the same length and 
tension as that of the belt ABFEDH, and what will be the ra- 
tio of the angular velocity of the two pulleys. 

Draw Ca, cb perpendicular to al, and from c draw cn per- 
pendicular to Ca; el and cn will be parallel to al, and hence 
will be = al: with the radius Cn describe a circle nrqst, and 
the tangent nc will be = the tangent al, = y(Cc?— (IC — hic)’). 

The arc rn is = arc la — arc Al. 

It is required from these data to find the diameter 7h. 

Scholium. In the solution of this problem, it will be meces- 
sary to express the arc rn in terms of its functions; the mode 
of conducting the solution will be left for the aiidentay 


7. Required an expression for the super- 
ficies, and also for the solidity of a tetraedron 
ABCD, in terms of its linear edge, AB=A. 

Ans. A’./3 = the surface. 
qzA’ V2 = the solidity. 


8. Required expressions for the surface J\ 
and solidity of a regular hexzedron or cube 
in terms of its edge. Ans. 6A’,= its surface. 

A*,= its solidity. 


238 QUESTIONS FOR EXERCISE. 


2 


9. How will you express the surface 
and solidity of an Octedron in terms of 
itsedge? Ans. 2A’*VW3 = the surface. 

sA° V2 = the solidity. 


10. Let the Surface and solidity of a do- 
decedron be expressed in terms of its 


edge. 
Ans. 15A’/(1+2./5)= the surface. 
47+21 7/5 
par = the solidity 


11. Express in terms of its edge, the sur- 
face and solidity of an Icoseedron. 


Ans. 5A’./3 = the surface, 
5A°/7+3/5 = the solidity. 
2 


ches, and its depth 8 inches: what must the diameter of a 
bushel be when its depth is 71 inches ? Ans. 19.1067. 
13. Of what diameter must the bore of a cannon be, which 
is cast for a ball of 24 lbs. weight, so that the diameter of the 
bore may be 1-10 of an inch more than that of the ball, and 
supposing a 9 lb. ball to measure 4 inches in diameter ? 
Ans. 5.646 inches. 
14. Suppose the ball on the top of St. Paul’s church is 6 
feet in diameter, what did the gilding of it cost, at 3d. per 
square inch ? Ans. £237, 10s. 1d. 
15. What will the gilding of a right rectangular revoloid, 
whose axis is 4 feet, come to at 5 cents per square inch ? 
Ans. $720. 
16. A silver cup, in form of the frustum of a cone, whose 
top diameter is 3 inches, its bottom diameter 4, and its altitude 
G6 inches, being filled with beer, a person drank out of it till he 
could see the middle of the bottom; it is required to find how 
much he drank ? | 
Ans. 42.899844 cubic inches = .152127 ale gallons, or 1 
gill and 4 nearly, the quantity required. 


QUESTIONS FOR EXERCISE. 239 


17. ‘Two persons would divide between them, by a plane 
perpendicular to the base, a hay rick, in the form of a para- 
boloid, whose altitude is 40, and the diameter of its base 30 
feet ; it is required to find the difference between the solidities 
of the parts, supposing the’ altitude of the section to be 28 
feet. Ans. 11265.758038, the difference required. 

18. In the construction ef a railroad, having contracted 
for the sum of $1000 to excavate acertain section, the area of 
whose conjugate:sections in 11 different places, taken at equal 
distances of 3 reds each, including the ends, are as follows, 
viz: the first, 150 square feet,the second, 160. the third, 165, 
the fourth, 172, the fifth, 190, the: sixth, 210, the seventh, 224, 
the eighth, 240, the ninth, 202, the tenth, 108, and the eleventh 
0. After having disposed of the materials to be excavated at 
10 cents per cubic yard, to be delivered on an adjoining «sec- 
tion, I afterward received an offer to have the whole labor of 
excavation and delivery performed for 30 cents per yard ; 
shall I gain or lose by my contract if I accept of the offer, 
and how much ? Ars. I shall gain $355,64. 

19. To determine the weight of a hollow spherical iron 
shell, 5 inches in diameter, the thickness of the metal being 
one inch ? Ans. 11.79]b. 

20. It is propesed to determine the proportional quantities 
of matter in the earth and moon; the density of the former 
being to that of the latter, as 10 to 7, and their diameters as 
7930 to 2160. Ans. as 71 to 1 nearly. 

21. ‘What difference is there, in point of weight, between 4 . 
~ block of marble containing 1 cubic foot and a half, and ano- 
ther of brass of the same dimensions, whose specific gravity 
is 8000? Ans. 496lb. 140z. 

22. What position in the line between the earth and moon, 
is their common centre of gravity ; supposing the earth’s di- 
ameter to be 7920 miles, and the moon’s 2160; also'the density 
of the former to that of the latter, as 99 to 68, or as 10 to 7 
nearly, and their mean distance 80 of the earth’s diameters? 

Ans. 633.65 miles below the surface of the earth. 

23. How deep will a cube of oak sink in common water ; 


each side of the cube being 1 foot ? Ans. 11,3; inches. 
24. How deep will a globe of oak sink in water; the dia- 
meter being | foot ? Ans. 9.9867 inches. 


25. If a cube of wood, floating in common water, have 
three inches of it dry above the water, and 4;$5 inches dry 
when in sea-water ; it is proposed to determine the magnitude 
of the cube, and what sort of wood it is made of ? 

Ans. the wood is oak, and each side 40 inches. 


240 QUESTIONS FOR EXERCISE. 


26. Hiero, king of Sicily, ordered his jeweller to make him 
a crown, containing 63 ounces of gold. The workmen 
thought that substituting part silver was only a proper perqui- 
site; but Hiero, suspecting that fraud had been practised, 
Archimides was appointed to examine it; who on putting it 
into a vessel of water, found it raised the fluid 8.2245 cubic 
inches ; and having discovered that the inch of gold more cri- 
tically weighed 10.86 ounces, he found by calculation what 
part of the king’s gold had been changed. And you are de- 
sired to repeat the process. Ans. 28.8 ounces. 

27. Supposing the cubic inch of common glass weigh: 
1.4921 ounces troy, the same of sea-water .59542, and of 
brandy .5368 ; then a seaman having a gallon of this liquor 
in a glass bottle, which weighs 3.84lb out of water, and, to 
conceal it from the officers of the customs, throws it over- 
board. It is proposed to determine, if it will sink, how much 
force will just buoy it up? Ans. 14.1496 ounces|] 

28. Suppose, by measurement, it be found that a man of 
war, with its ordinance, rigging, and appointments, sinks so 
deep as to displace 50000 cubic feet of fresh water ; what is 
the whole weight of the vessel ? Ans. 1395,, tons.. 

29. It is required to determine what would be the height of 
the atmosphere, if it were every where of the same density 
as at the surface of the earth, when: the quicksilver in the ba- 
rometer stands at 89 inches; and also, what would be the 
height of a water barometer at the same time? 

Ans. height of the air 291662 feet, or 5.5240 miles. 
height of water 35 feet. 

30. If the inner axis of a hollow globe of copper, exhausted 
of air, be 100 feet; what thickness must it be of, that it may 
just float in the air? Ans. .02624 of an inch thick. 

31. Ifa spherical baloon of copper, of ;4; of an ineh thick, 
have its cavity of 100 feet diameter, and be filled with inflam- 
mable air. of +4, of the gravity of common air, what weight 
will just balance it, and prevent it from rising up into the at- 
mosphere ? Ans. 21'785l|b. 

32. Construct a quantity == and show its value. (See 


Book IV., Chap. IL.) 


is 


33. I:xpress in a series of variables, the sym- 
etrical cylindrical ungulas, DBAECAB. 
where, AB=2I1D=z, and IC=z. 


QUESTIONS FOR EXERCISE. 241 


33. Required the solidity of the 
vacuity of a gothic roof, and also 
the solidity of the materials of the 
the roof ; the span AB = 30 feet, 
the chords of each are An and Bn 
=40 feet, the versed sine mo = 6 
feet, the thickness of the pear KA 
or BH = 17 feet, at the spring of 
the arc, the thickness of the crown ¥¥ 
of the arch Dn=4 feet of the roof 
60 feet. ™ 


( 52896.89319 cubic feet, the solidity 
er of the vacuity. 
" ) 104854.11776 cubic feet, the soli- 
i dity of the materials. 


34. Required the superficies of a dome in the form of a right 
hexagonal revoloid, each side of the base being 10 feet, and 
height 10 feet. 

Ans. 519.61524 square feet. 


35. The circumference of the base of a circular dome is 
150 feet, and its height 23.873 feet ; required the superficies. 
Ans. 3581.1 square feet. 


36. If the height BC of a saloon be 
3.2 feet, the BoD, of its front 4.5 feet, 
the distance or, of its middle part, 
from the arc 9 inches, and the mean 
circumference at m=50 feet ; required 
the solidity of the saloon. 

Ans. 138.26489 cubic feet. 


37. What is the whole surface of a saloon round a rectan- 
gular room, the mean compass at r=67.3187 feet, the girt DrB 
3.1416 feet, and the ceiling measures 16 feet long, and 12 feet 
broad ? 
Ans. 403.4727 square feet. 


38. Required the concave surface of a cireular arch, raised 
on arectangular base, whose sides are 27 feet 3 inches by 20 
feet 4 inches. | 
| Ans. 632.5415} square feet. 


242 DESCRIPTION OF A 


Description of an instrument constructed by the author, for 
measuring distances and heights, by a single observation, and 
without changing the position, or measuring any base line. 


The principle on which this instrument is constructed, and 
by which the result is produced, consists in arranging mirrors, 
_or reflectors, in such manner as to convey two distinct images 
of any distant object to the eye of the observer, as seen from 

two positions which are indicated by two mirrors, placed at 
any given distance from each other on the instrument, and 
causing the’ object to appear in two positions at the same time, 
and then measuring the apparent angle under which the two 
images appear. 

For this purpose the following diagram represents one form 
of the construction of the instrument. 


aa: 
|Z 
le 
AZ a 
| AGA x 
\|\GAA4 
ty 
AZ 
ge , g 1 i 
La 4a . 
h 
o> 
Bs 


AB, represents the stock or base; at the extremities of 
which let two mirrors, I, wu, be placed in the manner of the 
index glass to a common quadrant, and set perpendicular to 
the plane of the instrument, and at an angle of 45° with its 
axis or one of its edges; these we will call the object glasses, 
one of which, viz: 1, is placed on the centre of motion of an 
imdex, Mm, which is moveable about a centre at e, ‘and withit 
the mirror, which for distinction is called also the index glass. 
There are two other reflectors, n, 7. placed near the middle of 
the stock of the instrument, one above the other, their edges 
being in contact, the planes of which cross each other at right 
angles; one of these reflectors is parallel to one of the object 
glasses n, the other to the index glass, I. 

In using this instrument it must be placed or held so that 


TRIGONOMETRICAL INSTRUMENT. 243 


jts axis shall be perpendicular to a line from the object EF, to 
the fixed object glass u, and images of the object will be 
formed in the two object glasses I, K, and reflected into the 
two glasses n andi, and hence, to the eye of the observer, 
forming two distinct images, ef, ph, one of which appears 
higher than the other; and the angle under which the two ob- 
jects appear, varies according to the distance of the object. 
In order to measure the angle contained by the apparent po- 
sitions as indicated by the two images, the index is moved, and 
with it the index glass 7, till the object shall appear in the 
same position in both glasses, so that there would appear to 
‘be but one image formed in both, or till the images formed in 
both would appear identical. Nowif a vernier scale, v, is at- 
tached to the index and graduated, it will indicate the apparent 
angle under which the two images appear, or the angle under 
which the distance between the two object glasses would ap- 
pear if placed at the distance of the object. But by the law 
of reflection, the index would be moved only through half the 
angular distance of the two images in order to produce an ap- 
parent coincidence ; and hence the scale should be graduated 
with double the ordinary divisions, for the angle would be in- 
dicated by twice the angular motion of the index. In order 
that the index may be adjusted with accuracy, a tangent 
screw s,is provided,:by which it can be adjusted with any 
precision required. When the observation and adjustment is 
completed, we have a right angled triang'e whose base is the 
distance of the two mirrors, and whose altitude is the aistance 
of the fixed object glass to the object, and since we have one 
of the acute angles, the other side or distance of the object 
‘becomes known. 

When the object EF, is at any considerable distance, the 
‘angle uFI, or ull, becomes smaller, and the object will ap- 
spear to come to the mirror I, from the position GH ; so that 
HI and Fu produced, shall make an angle with each other 
equal to the apparent distance of the two images. But when 
the object is at an infinite distance, then the lines Fu, HI, be- 
come parallel, and the two images coincide, so’ that but one 
image appears to the observer; and hence, in this ‘case, the 
distance cannot be measured. 

In order to save the trouble of calculation in each case, a 
table may be constructed to accompany the instrument, which 
shall contain the distance corresponding to any given angle 
- pointed out on the scale, or, the scale itself may be graduated 
to specific distances, which may be read off instead of the 
angles. | 


Q44 DESCRIPTION OF A 


Another construction of the instrument, using but two 
glasses is as follows : 

Let two mirrors, a and }, be placed parallel to each other 
at the opposite extremities of the instrument, making an angle 
of 45° with its axis as before, and let the mirror }, be move- 
able about a centre by means of an index as described above. 


SCSOCC SRA  USRs Masseaasnye “+ SEO) ARS CORP SSSane 


° 
. 
® 
® 
° 
1 X 
o 
. 
4 
® 
‘ 


When any distance is to be measured by: this instrument, 18 
is placed so that a line from the object P, to the mirror a, shall 
be perpendicular to the axis 7m, of the instrument, the eye of 
the observer being at any point e, in the axis produced ; and 
the image of the object seen in the mirror a, will, by the law 
of reflection, appear in the direction of this axis ; and the im- 
age P, in the other mirror, if seen at all, will appear at 0, not 
coinciding with m. But by turning the mirror b about its axis, 
the image will advance toward v, in the line of the axis of the 
imstrument, and will ultimately coincide with it; whem the two 
images will be seen at e, in the same line, ev. 

It may be easily shown that the angular motion of the mir- 
ror b, necessary to bring it into the position b’ n’ so that the 
two images shall appear to coincide, will be half the angle at 
P, hence the distance is determined as before. 


TRIGONOMETRICAL INSTRUMENT. 245 


In this form of construction, I place the mirror in a wooden 
case or tube, leaving openings in its side in front of each mir- 
ror, and a small hole at the end for the eye of the observer ; 
the index and scale is on the out-side of the case: the mirror 
a, must occupy but half a section of the tube, so that the mir- 
ror ) may be seen over the edge of the mirror a; and the ob- 
ject is attained by bringing the iniages in both mirrors to co- 
incide, or so as to appear as one image. 

In order to determine the powers of this instrument, it is 
only necessary to observe that the distance mP, is the cotan- 
gent of the angle P to the radius mr, and that when mr is 
given, the value of mP may be calculated for any assumed 
value of the angle P, which is half the angle measured by the 
index and scale. Assuming the distance between the mirrors 
to be five feet, the angular motion of the mirror 6 from its po- 
‘sition parallel to a, the zero of the scale, will be for 1000 feet 
nearly 8'35”, and for 1100 feet '7'50”, a difference of 45” or 
three-fourths of a minute for a difference of 100 feet, or ten 
per cent of the first distance. 

If we assume that by the divisions on the scale attached to 
the index, the motion of the mirror may be correctly found to_ 
half minutes, then the. distance between the mirrors being 
taken at five feet, a change of half a minute would correspond 
at 10 feet to .007 of a foot, at 100 feet to .62 of a foot, at 
1000 feet to 63 feet, and at 10000 feet or 1.9 mile the whole 
angle is but 51”, and considerable variations would entirely 
escape detection; but by the application of a telescope to the 
mstrument in making the observations, its powers and accu- 
racy may be considerably extended. 

After having measured the distance to any object, as a 
house, or a tree, its altitude may be easily found by moving 
the index so that the top of the object in one mirror shall co- 
incidé with the bottom in the other, when the angle indicated 
on the scale, less the angle first found, corresponding to, the 
to the distance, is the angle under which the object appears ; 
whence having the angle and distance of the object, its alti- 
tude becomes known. 

The following investigation of the powers of the instru- 
ment, showing its limits of practical accuracy, is taken from 
a report furnished by a committee of the Franklin Institute of 
Pennsylvania, to whom the two instruments designated above 
were submitted by the author in 1833. (Published in Vol. 
XI., No. 8, Journal of the Franklin Institute.) 


246 DESCRIPTION OF A 


Call the variation in the angle P, y, the distance m P a, and 
mr, b. Suppose the angle P, to become P — y, and that then 
a+x, x denoting the increase of length of a, corresponding to 
a decrease, y, of the angle P. 

By trigonometry, 


fan,. Pb ys= Q and 
a 


tan. (P—y = Fee ar or by substituting for 


tan. P and tan. (P — y) their values found above. 
6 ® 
ds tan. y, 


GGhisc ov: cA aM» BHR 
+g” tan. y 


bs 2) — a> tan. h | 
atx a+b tan.y’ else 
joe (a Bibs be okt 
b—a X tan. y’ 
a’ +0° 
b 


eee eine Geen 


tan. ¥ 
If, as assumed above, b = 5, and y = 1, the general. equa- 
tion becomes 


i 


et 25+a 
v= 17241.4—a 
125 ¢ 
Wh 244 — Spa eS vi 
en a — 10,2 172314 007.- 
10.025 
Kor a= 100, z = aaa 


For a = 1000, xz = 61.6, and for a = 10,000, z = 13,809 ; 
which is greater than the distance a. 

By assuming a limit to the accuracy required, calculation 
will show how the instrument may be adapted to this limit 
when possible. Tor example, let the greatest inaccuracy al- 
lowed be one foot in 100, then b and y must be so adjusted 
that at the greatest distance for which the instrument is to be 


used z = a0" Calling this value of a, a’, we shall have, 


TRIGONOMETRICAL INSTRUMENT. 247 


qa? b? 
.O1 a’ = b sas or 
) —. q! 
tan. y 
O1 a’ 
ier re x 6 =— 1.01 a”, an equation which 


must exist in order that the required accuracy may be attain- 
able. ‘To examine by it the instrument already supposed, let 
us ascertain whether at 1000 feet, as the greatest distance at 
which it is to be used, the accuracy will come within the li- 
mit of one foot variation, in 100. In this case a’ = 1000, 8, 
as before, = 5, and tan. y = tan. 1’. Whence 2? = 25, 


= 17241.4, 0la' = 10, and 1.01 a? =1.010.000. Sub- 
tan. y 


stituting these values in the equation above, it requires 

25 — 172,414 = — 1,010,000, the equation is 
not fulfilled, and the instrument does not come up to the re- 
quirement. It would be easy to determine values of y and 6 
required for all possible degrees of accuracy, and thus by the 
possibility of making the half divisions accurate, and by the 
length which convenience might limit, to ascertain whether 
the instrument could be constructed to give the required de- 
gree of accuracy. 


: he B . a 
The investigation may be made more general, thus ; let 3 
express the required limit of accuracy at the greatest distance 
for which the instrument is to be used, then at that distance 


a g 
tiie HOT calling, as before, the value of a, a’, 


Gd ao 
o- thee rewnence 
n b 
tan.y 
/ 12 ] 
a pa MRE dy ta Wat 1 gti(T, hy 
nm tan. y Tes nN 
NAT’ AS AB FRO OH 
BITE tan. y 4 n* tan. TP Gi Oa ot 
1 
b=a(itV1- 4 n? tan. *y (1 + = 
2n tan. y 


This equation is possible when 4n tan. *y (n+1) < 1. 


Nata. 


a ee 


BOOK I. 


Havine in the preceding volume treated of the properties of the Paraboza, 
Ellipse, and Hyperbola, we show in this, that these curves are the sections 
of the cone, and are each formed by a plane passing through a cone, accord- 
ing to certain conditions. Of these sections the quadrature of the parabola 
iseasily attained, by the principles embraced in Prop. IV., in relation to the 
sectional divisions of a prism. Props. VI. and VII., show the application of 
the principles* to the quadrature of the parabola. 

The quadrature of the Milipse, though not so accurately determined as that 
of the parabola, is nevertheless easily expressed in terms of the circle’s 
quadrature. 

The Hyperbola is of more difficult determination, in relation to its quadra- 
ture, than the other conic sections, but, is nevertheless susceptible of being 
approximately determined to any extent required. 


BOOK II. 
ON SOLID SECTIONS OR SEGMENTS. 


Since the term section, though originally applied only to surfaces, made by 
cutting a solid, is extensively used in the arts, as expressing a definite por- 
tion of a solid, I have taken the liberty of making solid sections, as synoni- 
mous with segments of solids. This book, consists mostly of the compari- 


son of cylindric, and conical ungulas. 
BOOK IMI. 
OF REVOLOIDS. 


Revoloids are a class of bodies not usually treated of in works on Geom- 
etry ; but, from the important considerations connected with their organiza- 
tion, and from the relations which they bear, both to rectilineal and curvili- 
neal solids, it is of the highest importance to Geometry, that their properties 
should be discussed, and that they should receive a conspicuous place ~ 
among geometrical solids ; more especially, as they are almost the only cur- 
vilinear solids, that are absolutely cubable. For, we have shown, in the 
progress of the work, that, not only the right or spherical revoloid, and the 
elliptical revoloid, Props. VI. and VU. are cubable, is absolute terms, but 
also Parabolic and Hyperbolic Revoloids ; moreover, we have shown (Prop. 
Ill.,) the quadrature of the surface of a right revoloid, without regard to the 
circle’s quadrature, although it is bounded by cylindric surfaces. 

Some important principles are derived from the subject of mathematical 
transformation, as in Prop. IV., V., XI., and Corollaries. 


* In Prop. VI, the parallel lines AK &c., should be parallel to the axis EF of the parabola, 
instead of the position as there expressed, otherwise, the argument is not correct; but since the 
principle to be established, is not affected by the errorin that diagram and argument, it is 
deemed best not to alter the demonstration in this edition. The Scholium to Prop. VII, may 
be corrected by making AB or CD the axis of the parabola, instead of AC or BD. In all re- 
ferences to these propositions in the succeeding parts of the work, it will be perceived that the 
conditions of their application is expressed. ‘ 


a” 


NOTES. ' a 249 
BOOK IV. 


The quadrature of the revoloidal surface treated of in Book III., furnishes 
us with data for the quadrature of the circle, through the medium of the re- 
voloidal curve ; this subjectis amply discussed in this book ; through the 
properties of this curve, we are aiso enabled to deduce some important trig- 
nometrical functions. In Prop. Ifi., it is shown that the revoloidal curve 
may be by transformation derived from an elliptical curve. 

expressions are obtained in Props. IX. and XII., for the length of the cir- 
ele’s circumterence ; it is there shown that these approximations may be ex- 
tended indefinitely, so that the circumference may be obtained to any degree 
of exactness required. 

By applying the principles of the parabola to those of the revoloidal curve 
in Prop. XVIL., a remarkable approximation is obtained, so that if the sine 
and cosine of a small arc, and sine of half the given arc is obtained, the arc 
itself may be expressed in terms of those functions ; and it is shown that these 
functions may be so taken, that the arc shall be truly expressed to the same 
number of decimal places, that those functions are truly expressed ; in pur- 
suance of this, will be found, an example in Mensuration, page 194, where 
the are is correctly caleulated to 20 decimal places, by a simple process, 
having the sines and cosine given, as data, to 21 decimal places. Other 
modes of approximation, for the arc of the circumference may be pointed 
out, depending on the same principles, derived both from the quadrature of 
the revoloidal surface, and cubature of the revoloid; bat, by pursuing the 
course pointed out, page 125, the arc of the circumference may be found, 
with certainty, by this process, to any number of decimal places we have 
patience to pursue it. 

Props. X VIII and XIX, prepare us for the construction of a curve, described 
in Prop. XXI., termed the curve of the circle’s quadrature, the properties of 
which are, that a line drawn from any point in this curve, perpendicular to 
its conjugate diameter, will be equal to the are-of the inscribed circle cut off 
by a secant drawn from the centre of the circle to this point, and if another 
line be drawn from the same point in the curve, to the extremity of the 
conjugate diameter, the area of the space, intercepted by the two lines with- 
out the circle, will be equa! to the area of the segment of the circle cut off 
by the latter line, From these properties, we are enabled to deduce an im- 
portant theorem, in relation te segments of the circle, viz., that the area of 
any segment of a circle is equa] to the difference between the arc of the 
segment, and its sine, multiplied by half the radius. 


BOOK V. 


In this book, all geometrical magnitudes are discussed from their princi- 
ples of organization, from elementary magnitudes, without referring them to 
any specific forms or relations. And after introducing, and explaining the 
principles of the production of geometrical magnitudes, from variable ele- 
ments, in Chap. L., in order to render this science subject to analytical, and 
algebraic consideration, a peculiar notation has been introduced as the sub- 


ject of the second Chapter; and the mode of application, of this notation, 


to such subjects is there explained. 

By this mode of conducting geometrical investigations, and by this nota~ 
tion, we are enabled in a manner, somewhat more obvious than that of the 
calculus, to arrive at the same results as are obtained by that science, 

And although this forms but an introduction to the subject, yet i 
furnishes evidence of its adaptation to the investigation of the properties o 
all geometrical magnitudes : and may, perhaps, be rendered of equal univer 
sality with the calculus, with which it is most intimately allied, 


s 


250 NOTES. 


Here instead of the infinitely small momentary increments of variable mag- 
nitudes, or instead of the differentials of variables, this notation recognizes 
only the conditions of the variables themselves, in their associated capacity, 
which, from the principles of the science, may be integrated, as certainly, 
and with more obvious rationality, than those performed by the calculus from 
their differentials. 

By this notation, we are enabled to get a positive expression for the cir- 
cle’s quarature, in known functions of the diameter ; which, since all Geom- 
etricians are satisfied of the incommensurability of the circumference In 
direct terms of the diameter, should be received as the quadrature itself. 
For, from this expression, means may be devised of developing, decimally, 
the quadrature to any desirable extent. 

Chapter III: is an introduction to the differential and integral calculus, de- 
signed to show the first principles of that science, and to show its connec- 
tion with that of Chapter II., both of which are evidently in their essential 
particulars, based on the principles contained in Chapter I. 

Chapter IV. is devoted to the application of the. principles previously dis- 
cussed, to determine the position of the virtual centre, or centre of gravity of 
geometrical magnitudes. This has universally been denominated by all 
authors, hitherto as the centre of gravity. I doubt not, I shall have the ap- 
probation of most Mathematicians in discarding that term, and supplying in 
its place that of the virtual centre. The term centre of gravity, though per- 
fectly proper in works on Mechanics and Natural Philosophy, is highly 
incongruous in a work on pure mathematics, where the physical properties 


of bodies is not a subject of investigation, and perhaps not intelligibly un- 
derstood. 


MENSURATION. 


Such subjects in the mensuration of surfaces and solids, as could not con¢ 
sistently be introduced into the elementary part of the work, is introduced 
here; most of the subjects embraced in this, have been discussed in the geom- 
etrical part of the work, and the principles demonstrated ; where this is not 
the case, a reference has been made to the author, where such demonstra- 
tion may be found ; on this subject free use has been made of Hatton’s Men- 
suration ; limited quotations have also been made from other authors. 


The article on Guaging is mostly taken from Day’s Mathematics, though 
originally derived from Hutton. 


TABLE OF NATURAL SINES. 


In the following table of Natural Sines, the sines are exhibited to every de- 
gree and minute of the quadrant, and so arranged, that the degrees corres- 
ponding to the sines will be found on the top of the page in the same column 
with the sine ; the minutes in the left hand column opposite their sines: and 
the degrees answering to the cosine will be found in the same manner, at 
the bottom, with their minutes in the right hand column. 

The sine or cosine of an arc greater than 90°, is the same as the sine or 
cosine of its supplement; or the same as those of the difference of 180° 
and the given arc. 

The sines and cosines of any arcs greater than 180°, are the same as 
those for ares less than 186°, but with contrary signs, being by trigonometry 
considered negative ; they may be found in the table as above, by deducting 
180° from the given arcs. 

If the sine or cosine is required to degrees, minutes, and seconds. 

Take out the sines or cosines answering to the next less, and the next 
greater arc expressed in the table; multiply their difference by the given 
number of seconds and divide the product by 60; then the quotient added to 
the sine of the next less arc will be the sine or cosine required. 

Example. Required the sine of 32° 21' 45”, or its supplement 147° 38’ 15”. 


The sine of 32° 21’ is, .535090 
The sine of 32° 32’ is, 585335 
The difference is, .000245. 


Hence, 245 x 45+-60 
== .000245X 3 = .000184, the quantity to be added to .535090. 
Therefore, .535090 
+.000184 ° 


The sine required. = .535274. 

If the arc of a given sine or cosine is required in degrees, minutes, and 
seconds : 

Take out the arcs answering to the next less and the next greater sine 
and cosine, and multiply their difference by 60, and that product by the dif- 
ference of the next less, and the given sine or cosine divided by the differ- 
ence between the next less and next greater sine, and add or subtract as be. 
fore for the arc of the sine or cosine. 

Example. Required the degrees, minutes, and seconds corresponding to 
the sine .495994. 


The sine next less than that given is, -495964 
The next greater sine 1s, 496217 
The difference, .000253, 
The difference of the next less and the given sine is, 000030. 


The arc corresponding to the next less sine is 29° 44’, , 

Hence, 000030 x 60+-000253 = 7 = the number of seconds to add to 
29° 44’, 

Therefore, the required arc is 29° 44’ 7”, 


oh 


2 NATURAL SINES. ia 


m| 0° is Nl 8°.) Sb Re of OP fer? 'p 6? 9° |™ 
000000 017452 034899 052336069756 087 156|104528) 121869|139173/156434)6 - 
000291/017743'035 190/052626/070047 087446) 104818) 122158) 139461)156722'59 
000582|018034) 03548 1|05291 7070337 087735| 105107) 122447 139749'157009'58 
000873 018325 035772 053207070627 088025 | 105396) 122735) 140037|157296 57 
001164/018616/036062 053498 070917 088315} 105686) 123024! 140325|157584/56 
001454|018907 036353053788 071207 088605) 105975/ 123313 140613/157871/55 
001745/019197|036644/054079 071497 088894| 106264) 123601 140901/158158/54 
002036 019 188]036934 054369 071788 089 184/ 106553) 123890 141189|158445)53} 
002327 |019779}037225|054 660/07 2078 (89474 | 106843) 124179) 141477|158732)52} 
0026 18)020070) 137516054950 072368,089763} 107132) 124467, 141765) 15902059 
10,0029119|/029361| 137806 055241072658 090053) 10742 1| 124756) 142053) 159307/50 
11/0032U0| 020652! 138097055531) J72948 090343) 107710, 125045 149341 |159594 49 
1200349 10209421)38388]055822|073238/090633/ 107999 125333 143629/159881!48 
13/003782/92 1233 033678)056112'073528)090922'1 08289) 125622 142917/160168147 
14/004072/021524/038969 056402073818,091212) 10x57 125910/143205)160455/46 
15)004363/021815/039260,056693 074108 091502) 108867 126199/ 143403 160743/45 
16)004654/022106 039550)056983) 374399 091791)109156 126485/143780, 61030)44 
17/0049-45/022397/039841/057274/0746¢ 9 092081/109445) 126776) 144068 161317/43} 
18|005236 022687 040132 057564/074979 09237 1|109734/127065/144356/161604\42 
t9|005527|022978 040422 057854 075269) 092660] 110023| 127353144644) 161891141 
20/005818)023269 0407 13053145 075559 092950) 110313127642) 144932 162178|49 
21|006109/023560 041004/058435/075849 093239110602) 127930 145220162465 39 
22/006399/023851)/041294 058726 076139) 93529) 1 10891 )128219/145507|162752/38 
23 006690)024141/041585/059016 076429 093819) 111180) 128507|145795) 163039 37 
24/006931|924432/041876 059306/076719|J9410-|111469) 128796) 146083) 16332636 
25/007272\124723 0421 66\059597|077 009) 094395 |112758/129084/146371/163613 3A 
96|007563) 125014 042457/059887/077299/094687) 1 12047| 129373) 146659 163900/34 
27|007854} 125305 042748 060177/0775891094977) 112336/ 129661) 146946) 164787/33 
98|008145|025595|043938|060468/07 7879095267) 112625) 129949|147234/164474 32) 
29/008436) )25886) 043329 0607580781 69 095546) 1 12914)130238]147522)16476131} 
30008727 /)26177 043619 061049078459 095846) 113203 130526) 147809 165048 30 
31/0090 17} 326468)043910 061339 078749 096135/1 13492 130815/145097 165334 '29} 
32/0093.)8|026759 044201 064629)973039 096425} 113781 131103) 148385 165621 28 
33|009599| 127049)044491 061920079329 096714) 114070 131391) 148673165908 27 
34/00989)|027340/044782 062210 079619/097004| 114359) 131680 148960166195 26 
35,010181|27631|045072 062500/079909/097293) | 14648) 131968) 149248) 166452)25 
36|010472/027922|045363|06279 1/030 199/097583} 114937] 132256) 149535|166769)24 
37|010763|0282 12|045654/06308 1/080489 097872) 1 15226) 132545) 149823) 167056)23 
38)011054) 128503045944 063371 080779 098162 1 15515 /132833 150111 167342 22 
39|011344]02879 4 046235 063661/081069'09845 1/115804|133121 150398) 167629 21 
40/011635| 129085046525 063952081359 098741) 116093 133410|150686 167916 20/ 
41|01192+} )29375)04681 6) 64242/081649/09903U]1 16382) 133698 150973 16820319 
42\012217) 129666|947 106;064532)181939/099320)116671 133986 151261 168489 18 
43))12505)29957|047397) 064823 082228/099609) 116960) 134274/ 151548 168776 17) 
4.4/012799|)39218}947688) 065113 082518)099899) 117249134563 151836 169063 16) 
45/013990)| 80539/04797#|065403 082898) 100188) 117537) 134851) 152723 169350115 
46/013380)030829|048269\ 065693 083098) 100477} 127826) 135139|152411/169636 14 
47/01367 1) 131120/048559|065954 083383) 100767|118115,135427/ 152698 169923 13) 
48}013962|031 411|/048850,066274/083678) 101056)13844/135716 152986 170209 12 
49 ,014253)931702 049 140\066564 083965) 1013.46] 18693, 136004)153273 170490 11 
50,014544/ 131992/04943 1066854 84258 101635)/118982/136292)/153561 170783) 10 
51}014835|)32283 049721) 067145084547) 101924/119270 136580) 153848 171169) 9 
52\015126\032574 050012)067435)084837) 102214] 119559 136868) 154136 171356) 8 
53)015416 032864 050302) 067725085127) 102503!119848, 137156/154423 171643) 7 
54/015707/033155,059593) 068015 085417 102793) 120137 137445/154710,171929 6 
551015998|033146 050883) 068306085707) 103082/120426/137733 154998 172316) 5¢ 
56 016289) 033737 051174|068596 085997 10337 1)120714) 138021|155285 172502) 4 
57016580) 134027 051464) 068886 086286) 103661) 121003 138309 155572 172789) 3 
58016871] 134318 051755}069176 086576 103950) 121292 138597, 55860,173075| 2} 
i 

0 


COHAOMUEWNH SCS 


591017162) 134609 052045069466 086866) 104239) 121581) 138885) 56147) 173362 
a 34899 052336|069756 087156) 104528) 121869 139173) | 56434 173648 


89° | 88° | 87° | 86° | 85° | 84° | 83° | 82°} 81° | 80° |m 


Natural Co-sines. 


M 


173643 
173935 
174221 
(174508 
174794 
175080 
6|175367 
71175653 
81175939, 
9/176226 
10|176512 
11}176798 
12}177085 
13|177371 
14|177657 
15|177944 
16|178230 
178516 
178802 
179988 
179375) 
179661 
91179947 
180233 
41180519 
5/180805 
61181091 
181377 
181663 
29/181950 
30) 182236 
31) 182522 
32| 182808 
33183094 
34/183379 
35\183665 
36/183951 
37| 184237, 
38) 184523, 
39) 184899) 
40}185095 
41/1853: 
42185667 
43185952 
44)186238 
451185524 
46/186810 


om ONO re Cc 


19@8u9 
191095 
191380 
191666 
191951 
192237 
192522 
192807 
193093 
193370 
193664 


m| 10° | 11° | 12° 


193949 


NATURAL SINES. 


13° 


207912) 224951 
208196/225234 
208480) 225518 
208765) 225801 
20950) 226085 
29334) 226368 
209619|226651 
209903) 226935 
210187/227218 
210472)/227501 
210756) 227784 


211040/228068 


| 


14° 


241922) 


242204 
242486 
242769 
243051 
243333 
243615 
243897 
244179 
244461 
244743 
245025 


15° 
252819 
259 100 
259381 
259662 
259943 
260224 
260505 
260785 
261066 
261347 
261628 
261908 


Bic 
292372 
292650 
292928 
2)3296 
293484 
277035)293762 
277315/294040 
277594/294318 
277874)291596 
278153/294874 
278432|295152 
278712|295430 


16° 
275637, 
275917 
276197 
276476 
276756 


| 


18° 
309017 
309294 
309570 
309847 
310123 
310400 
310676/327218 
310953/327493 
311229|32776x 
311506/328042 


19° 
425568 
325843 
326118 
326393 
326668 
326943 


311782)/328317 


M 
60 
59 
58 
o7 
56 
95 
54 
53 
52 
5] 
30 


312059 


194234/211325 
194520/211609 
194805/211893 
195990/212178 
195376)212462 
195661/212746 
195946/213930 
196231/213315 
196517/213599 
196802)213 83 
L97087|214167 
197372/214451 
197657/214735 
197942)215919 
198228) 215303 
198513)/215588 
198798)215872 
199983'216156 
19936x%)/ 216440 
199653)2 16724 
199938/217008; 
200223) 217292 
200598)217575 
200793)217859 
201078)218143 
201363)218427 
2)1648/218711 
201933218995 
202218)219279 
202502/219562 
202787/2 19846 
2113 972)22013) 
293357/220414 
203642/220697 


228351 
22863 4 
223917] 
229200 
229484 
299767 
230050 
230333 
230616 
230899 
231182 
231465 
231748 
232031 
232314 
932597 
232880 
233163 
233 145250330 
23372“| 250662 
2340111250943 
2342941251225 
234577/25 1506 
234859/251788 
935142)252069 
9354251252351 
935708252632 
2359901252914 
236273}253195 
9365561253477 
2368381253758 
937121/254039 
237403/254321 
237686/254602 


245307 
245589 
215871 
246153 
246435 
246717 
246999 
247281 
247563 
247845 
248126 
248408 
2148690 
248972 
249253 
249535 
249817 


| 


250098; 


203927 


47 
48 
49 
50 
5] 
52 
53 
54 
55 
56 
o7 
58 
59 
60 


M 


3 


187996) 294211 
187381 
187667 
187953 
188233 
188524 
188-10 
189095 
189331 


20478] 
25065 
205350 
295635 
25920 
206204 
206489 
189667 206773 
189952 207058 
190238207343 
190523 207627 


24496, 


190809 207912 
oer 18° 


22098 1/237968; 
221265)238251 
221548)/2338533 
221832)/238316 
222116/239098 
222399 /239381 
222683 239663 
222967/239946 
223250 
223534 
223317 
224101 
224384 
224668241649), 
22495 1/241922 


77° | 76° | 


240793 


254883) 


255165 
255446 
255727 
256008 
256289 
25657 1 
256852 


257695 


252538 
258819 


————$__— 


15° 


262189 
262470 
262751 
253931 
263392 
263592 
2633873 
264154 
261434 
264715 
264095 
(5276 
265556 
255837 
266117 
26397/283179 
266678) 283457 
266958/283736 
267232) 284015 
267519) 284294 
267799/284573 
268079) 284852 
268359/285131 
268640) 25410 
268920) 285688 
269200285967 
269480) 286246 
269760/286525 
270040) 286803 
270320) 287082 
270600\287361 
270880287639 
271160/28791& 
271440/2838196 
271720/288475 
272000) 288753 
|272280/289032 
272560)289310 
(272841129589; 
273 120/269-67 


278991 
279270 
279550 
279829 
270108 
23038* 
280667 
230946 
281225 
231504 
231783 
282062 
282341 
282620 
282900 


295708 
295986 
296264 
296542 


312335 
312611 
312888 
313064 
296819)313446 
297097/313716 
297375/313992 
297653314269 
297930/314545 
298208/3 14521 
298486/315097 
2987631315373 
2990411315649 
299318315925 
299596316201 
299873 316477 
300151/316753 
300428 317029 
[300706 317305 
300983 317580 
'301261/317856 
'391538|318132 
301815,318408 
302093'3 18684 
302370318959 
'392647/319225 
392924319511 
303202/319786 
3034791320062 
3 375€1320337 
304033/320613 
30431: 320889 
304587 


321164 
304864 


321439 
305141/321715 
30541 }321999 
305695) 322266 
305972) 322541 
306249)322816 
306526323 192 


273400)290145, 
273679|290424 


249228 257133 273959/290702 307357 
'240510/257414|274239/290981/307433 


274519) 291259 


241075 257976 27479829 1537 308187/324743, 
241357 258257 275078|29 1815 308464/325018 


275358) 292994 
275637292372 


74° |. 73° | 


Natural 


Co-sines. 


306803/323367 
3970811323642) 
323917 
324193 
3)7910/324468 


308740/325293 


328592149 
32%867 48 
329141147 
329416146 
329691145 
329965144 
330240143 
330514/42} 
330789 41 
331063140 
33133839 
331612/38 
331887/37 
332161136 
332435135 
332710,34 
332984/33 
333258)32 
333533131 
333807130 
334081/29 
334355128 
334629 27 
334903126 
335178125 
33545224 
335726123 
335000129 
33627421 
336547|20 
‘33682119 
33709518 
337369{17 
337643116 
337917/15 
328190114 
238464113 
338738112 
3390 12/11 
339285110 
339559] 9 
339832) 8 
340106] 7 
340380] 6 
340653) 5 
340927] 4 
341200] 3 
341473] 2 
341747] 1 


319017|325568 


a a 


me TI" 


342020) 0 


70° |M 


4 


m| 20° 


~0'342020 


11342293 


2'349567 


31342340 
4'343113 
5'343387 
6343660 
7/343933 
81344206 
9344479, 
101344752 


111345025/361353/377571|393675 


121345298 
131345571 
14/345844 
15346117 
16/346390 
17/346663 
18|346936 
191347298 
90'317481 
91|347754 
991348027 
931348999 
241318572 
25)348845 
261349117 
971349390 
981349662 
99|349935 
301350297 
39'350752 
331351025 
341351297) 
351351569 
3613518 12 
37/352114 
381352386 
391352658 
401352931 
41135323 
421353175 
43353747 
41354019 
151354291 
46354563 
47354335 
48 
49 
50) 


353379 
3554651 


511355923 372078 


a2 


358363/374607 
358640)374876 
358911/375146 
359183/375416 
3594541375685 
359725/375955 
359997/376224 
360268/376494 
360540/376763; 
360811/377033 


| 


Zio Pggetr 23° 


NATURAL SINES. 


390731 
39099) 


392070 
392337 
392605 
392872 
393140 


361082)377302 


361625)/377841 
361896/378110 
362167/378379 
362433/378649 
362709/378918 
362980/379187 
36325 1/379456 
35352 2)379725 
363793/379994 
364064)330253 
364335/380532 
354606/380801 
364877/331070 
365148)331339) 
365418/381608 


393407 


393942 
394299 
394477 
394744 
395011 
395278 
395546 
395813 
396080 
396347 
396614 
396881 
397148 
397415 
397682 


391267 
391534 
391802) 


4- 

\4U06737 
407002 
407268 
407534 
407799 
408065 
408330 
408596 
408861 
409127 
499392 
409658 
409923 
410188 
410454 
410719 
410984/426832 
411249}427095 
411514/427358 
411779}427621 
412015/427884 
412310/428147 
412575/428410 
4.12849)428672 
413104)4298935 
413359/429198 
413634/429461 


25° 
422618 
422882 
423145 
423409 
423673 
423936 
424199 
424463 
424726 
424990 
425253 
425516 
425779 
426042 
426306 
126569 


26° 
(438371 
(438633 
433894 
439155 


A539 10 
454250 
454509 
| (454768 
439417/455027 
439678)455286 
439939)455545 
44()206}455804 
440462/456063 
440723) 156322 


441245}456339 
441506) 457698 
441767/457357 
442028457615 
442289/457874 
442550/458133 
442810)45839 | 
443071/458659 
443332/458908 
443593/ 459166 
443853 459425 
444114'459683; 
144375)459942 
444635/460200 
144896)460458, 
445156460716 


Oa i 


440984) 4156580) 


28° 
469472 
469728 
469985 
470242 
470499 
AT0755 


29° 
484810 
485064 
\485318 


485827 
486081 


486590 
486844 
487098 
487352 
487606 
487860 
488114 
488367 
488621 
488875 
489129 
489382 
489636 
489890 
490143 
490397 
490659 
490904 
AQ1157 
491411 


471268 
AT1525 
471782 
472038 


472294 
472551 

472807 
473063 
473320 
473576 
473832 
474086 
4743.44 
474600 
474356 
475112 
475368 
475624 
475880) 
176136 


485573 


471012'486335 


355.107 


365689 
365960 
366231 

366501 

366772 
367042 
367313 
367584 
367854 
368125 
368395 
368665 
368936 
36926 
359476 
369747 
370017 
370287 
370557 
379828 
371098 
371358 
371632 
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53 
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35619 1372448 
356486 372718 
356733 372988 
357010373258 
357281 373528 
357553 373797 


398096 374337 


352368 374607 


357825 374067 


381877/397949 
382146 398215 
382415 398482 
382683 398749 
332952/399016 
38322 1}399283 
333490)399549 
333758 | 399816 
334027 400082 
334295 400319 


413899 
414164 
414429 
414693 


429723) 445417 
429986 | 445677 
430249} 145937 
130511|/446198 


460974'476392 
461232476647 
461491 476903 
461749 477159) 


414950 


415487 
415752 


416281 


33456 4490606 
334332 400832 
385101 49 t149 
335369) 491415 
335633)/4016381 
335906 401948 
336174/402214 
386443 4024280! 
386711/402747 
336979) 403013 
387247 /403279 
337516493545 
337784/403811 
388052'404078 
388320 404344 
388588/401610 
388856) 404876 
3891241495142 
389392\405408 
389560 405673 
389928) 495939 
39) 196) 406205 
390 163 406470 
399731406737 


66° 


116545 
416810) 
417074 

417338 
417603 
417867 
41813! 

418396 
418650 
418924 
41918~ 
419452 
419716 
419980 


420244 


415223) 43 1036/446718) 


431561)447239 
416016/431823| 447499 


130774/446458 
431295446979 


4329861147759 
132348) 148019 
432610) 448279 
432873) 448539 
433 135|448799 
433397) 449059 
433659) 149319) 
433921|449579 


462007 477414 
462265 477670 
462523 477925 
462780/478181 
4630381478436 
463296/478592 
463554 478947 
163812|479203 
46469|479458) 
464327|179713 
464504/479968 
464849) 480223 
465100) 480479 


434183/449839 
434445/459098 
434707|450358 
434969] 150618 
135231/450878 
435493451137) 
435755)451397, 


#2601 7/451656 


465357)/480734 
465615)489939 
465872)4812 44 
466129/481499 
466387/4*1754 
465644) 482009 
46690 1)482263, 


4671 5}482518 


491664 
491917 
492170 


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60 
59 
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43 
42 
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39 
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37 
36 
35 
34 
33 
32 
31 


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492677 
492939 
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493436 
493689 
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494448 
494700 
494953 
495206 
495459 

495711 

495964 
496217 

496469 
496722 
496974 
497226 
497479 


497731| 9 


420508 
420772 
421636 
421300 
491563 
421827 
422191 
422355 


436278 
136540 
436802 
437963 
137325 
4373587 
A3 73848 
138119 


| 
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| 


151916467416 
452175 467673 
45243546793" 
152694 468187 
152953 468444 
453213 468701 
153472 468958 
153731469215 


422618 


438371 
| 65° 


64° 


153999) 469472 
63° | 62° 


482773497983 
483028) 498236 
483292)498488 
483537|498740 
483792/498992 
484046/499244 
484391/499496 
484555/499748 
4848 10/590000 


30) 
29 
28} 
27} 
26 
23 
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23 
22 
21 
20) 
19 
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17 
16 
15 
14; 
13 
12 
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56 
155 
O4 
52 
5] 
50 
49} 
48 
47} 
46 
45 
44 


i hh“ 7 


I}Zlomwmannarme 


Natural Co-sines. 


NATURAL SINES. 5 


a; 32 

529919 
530166 
530413 
530659 
530906 
531153 
531399 


m! 30° 


0500000 515035 
1)500252 515237 
2500504/515537 
500756 515786 
501007 516035 
501259 516284 
501511516533 
7/591762 516782 
8502014517031 
9/502266 517280 
10'502517/517529 


11/502769 517778 


34° 


{| —_—____— 
(559193 
599434 
5999675 
559916 
560157 
560398 
(960639 
560880 
561121 
561361 
561602 


261843 


33° 
544639 
544883 
545127 
945371 
945615 
545858 
d46102 
331645) 546346 
53189 1)546589 
532138) 546833 
532334/547076 


53263.:|547320 


35°"| 36° 

5735761587785 
573815\588021 
574053588256 
574291588491 
574529 588726 
574767|/588961 
575005'589196 


37° | 38° 
601815)615661 
602047/615891 
602280/616120 
602512616349 
602744'616578 
602976,616897|630450/55 
603208) 617036|630676)54 
575243'589431/603440)617265|630902'53 
97548 1/589666|603672/617494)/631127/52 
575719 589901 /603904/617722)631353)/5] 
575957/|590136/604136/617951/631578/50 


576195|/590371|604367/618180/631804/49 


39° lu 
629320160 
629546159 


12}503020 
13)503271 
503523 
503774 
504025) 
504276 
50452> 
O04779) 
50503 
DV523 

505532 
505783 
4506034 


518027 | 532876 
518276/533122 
518525)533365 
518773/533615 
519022533367 
519271534106 
519519)534352 
519768) 53459> 
529016/5348 44 
320205)535090 
5295 13/535335 
520761/535581 
521010)535827 
521258)536072 


28|507037 
29/597288 
30/507538 
31)507789 
32508049 
33/598290 
34/508541 
35}508791 


38/509542 
39}509792 


5|521506 


3) 525224539996 


536318 
521754/536563 
522002|536809 
522251537054 
522499|537300 
522747|537545 
5229951537790 
523242|533035 
523.490|538281 
523738) 533526 
523986|53877| 
521234|539016 
524481/539261 
524729|539506 
524977539751 


525472/540240 


43/510793 
44/511043 
45|511293 
46/511543 
47/511793 
48/512043 
49/512293 
50/512543 
51512792 
52/513 42 
53/5 13292 
54/513541 


58514539 
59/514789 
60515038 


Mm!) 59° 


525719)540485 
525967/540730 
526214)540974 
526461/541219 
526709) 541464 
526956/541708 
527293)541953 
527450|542197 
52769 7542442 
527944)542686 
52819 1)542930; 
528433|543174 
528685543419) 
(528932)543663 
529179|543907 
529426544151 


347563 
3947807 
348050 
348293 
948536 
548780 
549023 
549266 
549599 
549752 
949995 
550238 
550481 
550724 
550966 
551209 
551452 


562083 
562324 
562564 
562805 
563045 
563286 
563526 
563766 
564907 
564247 
564487 
564727 
564967 
565207 
565447 


576432590606 
976670/590840 
976908|59 L075 
577145/591310 
577383/591544 
577620|591779 
977858)592013 
578095 )592248 
578332/592482 
978570592716 
578807 |592951 
579044/593185 
579281 |593419 
979518)593653 
579755|593887 


565687 
5965927 


551694 


552422 
552664 
592907 
593149 
553392 
553634 
593376 
554118 
504357 
954602 
554844 
555086 
595328 
5999970 


557262 
597904 
557745 
557987 
558228 
558469 
5587 10 


529673544395 
(5299 19|544639 


“58° | 57° | 


558952 
559193 


56° 


566166 


565886 
967125 
567365 
567604 
5967844 
568083 
568323 
568562 
568801 


569I40 
5692380 
569519 
599758 


579992)594121 
580229/594355 
580466] 594589 


551937|566406/580703)594823 
99218: |566646/589941)/595057 
581176|595290) 609223 622970, 
581413/595524/609454 
581650)595758) 609684 
581886/595991)609915 
582123'596225/610145 
582359/596458)6 LO3TE 
58259 €|596692'6 10696 
59€925)\610836 


582832 
523069|597 159 
583305|5973I2 
583541)597625 
5£3777/597858 
584014)598992 


569997|584250/ 598325 


571997 
572146 
572384 
572623 
572261 
573100 
573338 
573576 


5D: 


535812)570236/58442/598558 
558054/570475/584722/59879 | 
556296)/5707 14|584958/ 599024 
556537/570952)585194/599256 
556779/571 191)585429|599 489 


59702 '|57 1430/585665|599722 
571669!585901/599955 


586137/600188 
586372/ £09420 
586608) 690653 
58684 4/600885 
587079] i01118 
58731 4|501359 


587550601583 
587785691815 


54° | 53° 


Natural Co-sines. 


618408 
618637 
618865 
619094 
619322 
619551 
619779 
620007 
620235 
620464 
620692 
620920 


604599 
604831 
605062 
605294 
605526 
605757 
605988 
606220 
606451 
606682 
606914 
607145 
607376 
607607 
607838|621604 
608069|621831 
608300|622059 
608531 622287 
608761 622515 
608992 622742 


621376 


623197 
623425 
(23652 


624107 
624334 
624561 


611067|624789 


611297 


'611757/625470 
611987|625697 
612217/625923 
612447/626150 
§12677| 626377 
612907|626604 
613137|626830 
613367|/627057 
(613596) 627284 
(613826 
614056)627737 
§14285)627963 
614515/628189 
614744|628416 
614974| 628642 
6152)3) 628868 
61543 2}629094 
615661)/629320 


5H. | 51° 


621148) 


623880) 6¢ 


625016 6: 19 
611527/625243. 


627510; 


632029 
632255 
632480 
632705 
632931 
633156 
633381 
633606141 
633831!40} 
634056|39 
634281138 
634506137 
634731136} 
634955135 
635180|34 
635405/33 
635629/32 
635854131 
63607830 
63630329 
636527/28] 
636751127} 
63976126 
637200125 


48 
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46 
45 
44 
43 
42 


638768)18 
638992117 
639215)16 
639439]15 
639663|14] 
63988613} 
640110/12 
64033311 
640557|10 
64078) 
641003 
641226 
£41450 
611673 
641896 
642119 
642342 
642565 
642788 


50° 


Sto es tn tone 


6 


M 


SONIA wWNWrH CS 


28 
99 
30 
31 
32) 
33 


37 


Ti 645236 


911647455 


41/651878 


511654081 


40° 
642788 
643010 
643233 
643456, 
643679 
643901 
644124 
644346 
644569 
644791 


645013 


645458 
645680) 
645902) 
646124 
646346 
646568 
546790 
f47012 
647233 


647677 
647898 
648120 
648341 
§48563 
648784 
649006 
649227 
649448 
649669 
649890 
50111 


650995 
651216 
651437 
651657 


65298 
652319 
652539 
652760 
652989 


653361 


654301 
654521 
654741 
654961 
655180 
655400 
655620 
655839 


656059 


656059 
656279 
656498 
656717 
656937 
657156 
657375 
657594 
657314 
658033 


653471 
658689 
658908 
659127 
659346 
659565 
659783 
660002 
660220 
660439 
660657 
660875 
661094 
661312 
661530 
561748 
661966 
662184 
662402 
662620 
662833 
663056 
663273 
663491 
663709 
663926 
664144 
664361 
664579 
664796 


665013 
665230 
665448 
665665 
665882 
666099 


42° 
669131 
669347 
669563 
69779 
669995 
670211 
670427 
670642 
670858 
671074 
671289 


671505 


41° 


558252 


671721 
671936 
672151 
672367 
672582 
672797 
673013 
673228 
673443 
673658 
673873 
674088 
674302 
674517 
674732 
674947 
675161 
675376 
675590 
675805 
676919 
676233 
676448 
676662 
676876 
677090 
677304 
677518 
677732 
677946 
678160 
678373 
678587 
678801 
679014 
679228 
679441 
679654 
679868 


680081 
680295 


666316 
666532 
666749 
666966 
667183 
667399 
667616 
667833 
668049 
662265 
668482 
668698 
6689 14/681786 
669131)681998 


680721 
680934 
681147 
681360 
681573 


680508) 


NATURAL SINES. 


yd 

681998 
§82211 
82424 
682636 
682849 
683061 
683274 
683486 
633598 
683911 
684123 
684335 
684547 
§84759 
684971 

685183 
§85395 
685607 

685818 
686039 
686242 
686453 
686665 
686876 
687088 
687299 
687510 
687721 

687932 
§88144 
688355 
6388566 
688776 
688987 

689198 
689409 
689620 
689830 
690041 

690251 

690462 
699672 
690882 
691093 

691303 

691513 

691723 
691933 

692143 
692353 
692563, 
692773 
692983) 
693192 
693102) 
693611 
693821 

694039 
694240 
694449 
694658 


48° | 47° 


46° 


44° 
694658 
694868 
695077 
695286 
695495 
695704 
695913 
696122 
696330 
696539 
696748 


636957 


697165 
697374 
697582 
697790 
697999 
698207 
698415 
698623 
698832 


69940 
699248 
699455 
699663 
699871 
700079 
700287 
70049 4 
700702 
700909 


TOLLL7 


701324 
701531 
701739 
701946 
702153 
702360 
702567 
702774 
792981 


TO3 L828 
703395 


703601 


703808 
TOAD15 
704221 


704428 
704634 
704841 
705047 
705253 
705459 
705665 
705872 
706978 
706284 
706489 
706695 
706901 
707107 


45° 


45° 
707107 
707312 
707518 
707723 
707929 
708134 
708340 
708545 
708750 
708956 
709161 


709366 
709571 
709776 
709981 
710185 
710390 
710595 
710799 
711004 
711209 


711413 
711617 
711822 
712026 
712230 
712434 
712639 
712843 
713047 
713250 
713454 
713658 
713862 
714066 


46° 


719340} 


719542 
719744 
719946 
720148 
720349 
720551 
720753 
720954 
721156 
721357 
721559 
721760 
721962 
722163 
722364 
722565 
722766 
722967 
723168 
723369 
723570 
723771 
723971 
724172 
724372 
724573 
724773 
724974 
725174 
725374 
725575 
725775 
725975 
726175 


714269 
714473 
714676 
714880 
7150-3 
715236 
715495 
715693 
715896 
716999 
716302 
716505 
716708 
716911 
717113 
717316 


717519 
717721 
717924 
718126 
718329 
718531 
718733 
71893° 


T1913 
719340 


726375 
726575 
726775 
726975 
727174 
727374 
727573 
727773 
727972 
728172 
72837 | 
728570 
728769 
728969 
729168 
729367 
729566: 
729765 
729963 
730162 
73361 
730560 
730758 
73957 
731155 
731354 


48° 


743145 
743339 
743534 
743728 
743923 
744117 
744312 
744506 
744700 


47° 
731354 
731552 
731750 
731949 
732147 
732345 
732543 
732741 
732939 
733137 
733334 
733532 
733730 
733927 
734125 
734323 
734520 
734717 
734915 
735112 
735309 
735506 
735703 
735900 
736097 
736294 
736491 
736687 
736884 
737081/748763 
737277 748956 
737474749 148 
737670 749341 
737867 749534) 
738063 749726 
7322591749919 
738455|75011 1 
738651|750303, 
738848|750496 
739043'750688 
739239] 750880 


739435|75 L072 
7396311751264 
739827|751456 
740023/751648; 
740218/751840 
749414)752032, 
740609) 752223 
740805| 752415 
741000/752606 
741195}/752798 
741391}752989 
741586/753181 
741781 
741976 
742171 
742356 
742561 
742755 
742950 


745088 
745282 
745476 
745670) 
745864 
746057 
746251 
746445 
746638 
746832 
747025 
747218 
TA7T412 
747605 
747798 
74799 1 
748184 
748377 
748570 


753563 
753755 
753946 
754137 
754328 
754519 


744894 756425 


753372 


49° 
794710 
754900 
755091 
755282 
‘755472 
755663 
755853 
756044 
756234 


M 
60 
59 
98 
o7 
26 
by 
54 
53 
52 
3] 
50 
49 
48 
47 
46 
45 
44 
43 
42 
4] 
40 
39 
38 
37 
36 
35 
34 
33 
32 
31 
30): 
29 
28 
27). 
26 
25]. 
24 
23 


756615 
756805 
796995 
797185 
757375 
797565 
T57755 
T57945 
758134 
758324 
758514 


758703 
758893 
759082 
759271 
759461 
759650 
759839 
769028 
760217 
760406 
760595 
760784 
760972 
761161 
761350 
761538 
761727 
761915|22 
762104/21 
762292/20 
762480)19 
762668) 18 
762856) 17 
763044) 16 
763232|15 
763420) 14 
763608)13 
763796) 12 
763984) 11 
764171/10 
764359 
764547 
764734 
764921 
765109 
765296 
765483 
765670 
765857 


743145)754710 


766044 


44° 


43° 


42° | 41° 


Natural Co-sines. 


40° 


Sol aorta sat ota Nee 


NATURAL SINES. ott ae 


m| 50° | 51° | 52° | 58° | 54° | 


0} 766044/777146] 78801 1}798636 809017 /819152/829038)83867 1|84848'857 167 |60 
1)766231)777329) 788190) 798811809188 /819319/829200/838829 848202'857317 159 
2|766418) 777512! 788369) 798985/809359 |819486 8293 63|838987|848356/857467 58} 
3/766605/ 777695 | 788548) 799 160/809530 |819652/829525/839146/848510/857616157 
4'766792 777778) 788727 | 799335'809700 819819/829688/839304| 848664857766 156 
5|766979/ 778060) 788905|799510/809871 |819985/829850|839462 848818857915 155 
6 
7 
8 


55° | 56° | 57° | 58° | 59° |m 


—_—_ 


767165) 778243 | 789084 | 799685 /810042 |820152/830012|339620/848972 |858065 154 
767352 778426) 789263] 799859|810212 |820318/830174|339778)849 125 |858214 153 
767538) 778608) 789441 |300034'810383 |820485 830337 |339936'849279 858364159 
9|767725) 778791 | 789620 |3800208)/810553 820651 |830499 | 340094 849433 |858513/5] 
10|767911,778973) 78979¢ |800383 810723 |820817/83066] | 340251/849586/ 3858662 154 
11/768097|779156| 789977 | 30055 /|810894.|820983/83082° | 340409 /849739/858811 |49 
12) 768284779338) 790155|/300731/311064/821149|830984!340567/849893 358960 |48 
13|768470\779520} 790333 |300906/811234 |821315/831146|840724'850046|359109 47 
14/768656'779702) 790511) 301080/811404 821481 |831308/840882!/850199|359258 | 46 
15|768842 779884) 790690 |/801254/311574 |821647|831470/841039/850352 359406 |45 
16) 769023) 780067) 790868 /3801428/311744 |821813/831631/841196)/850505 |859555 |44 
17/769214 780249) 791046/801602/311914 |821978/831793/841354/850658 |859704 |43 
18) 769400) 780430) 791224/301776)/312084|822144/831954/841511/850811)/859852 |49 
19|769585 780612) 79 1401/801949|312253 |822310)/832115/841668)850964/860001 |4) 
20|769771)/780794|791579/802123)/312423 |822475 /832277/841825) 851117 |860149 |49 


21)769957|780976)791757|80229 7| 312592 |82264.1 /832438/841982/851269 |860297 139 
22)770142!781157|791935/802470/312762 |822806|332599 |842139/851422|860446 |38 
23/770328|781339)792112/802644) 312931 |822971/332760|842296)851575|860594 |37 
24/770513|781520|792290)802817/313101 |823136/332921|842452/851727 |860742 |36 
25/770699/781 702) 792467/80299 1 /313270 |823302)|333082/842609/ 851879860890 135 
26/770884| 781883) 792644|803 164) 313439 |823467 |333243/842766)852032/861038 |34 
27/771069|782065| 792822) 803337 | 313608 |823632/333404)/842922 852184 /861186 |33 
28| 771254! 782246) 792999)/803511/313778 |823797 | 333565843079 852336 /861334 |32 
29771440 782427| 793176 803684313947 |823961/333725/843235 852488/861481 |3t 
30/771625!782608)793353/803857| 314116 |824126/333886/843391 852640861629 |30 


311771810 782789|793530 804030'314284/82429 1/834046 843548 852792 861777 }29 
32)771995 782970| 793707 804203 / 814453 |824456/834207 843704 852944|861924 28 
33'772179 783151|793884/804376/814622 |824620/334367|/843860 853096 862072 27 
34|772364|783332)794061)/804548) 81479 1 |824785/334527|844016)853248 862219 |26 
35|772549 | 7835 13|794238/804721/814959 |824949|334688 8441 72/853399 862366 |25 
36/772734'783693|794415|804894/815128/825113/334848 844328 853551 |862514 |24 
37|772918)783874) 79459 1|805066|815296 (825278 335008/844484'853702 862661 |23 
38/773 103 784055) 794768) 805239 /815465 /825442/335168/844640/853854 862808 |29 
39|773287/784235/ 794944 80541 1/815633 |825606/335328)/844795/854005|862955 |21 
40|773472) 784416)795121/805584/815801 |825770| 335488844951 /854156/863102 |20 


41/773656 784596] 795297 |805756/815969 /825934/335648 845106/854308 863249 |19 
42)773840| 784776) 795473 |805928/816138 |826098)835807/845262/854459 863396 |18 
43 774024 784957|795650/806100/816306 826262) 335967/845417/854610 863542 |17 
44|774209)/785137|795826 806273'816474 |82642€/336127/845573/854761|863689 | 16 
45|774393)/785317|796002 806445 816642 |826590)336286/845728)854912/863836 |15 
46774577) 785497|796178)|806617/816809 |826753) 33641 6/845883)/855063 [863982 | 14 
47'774761'785677|796354/806788 816977 |826917/336605/846038/855214 864128 |13 
48)774944 785857|796530 806960817145 |827081) 336764 846193/855364 864275 |12 
49|775128/786037|796706/807132/817313 827244] 336924/846348) 855515 86442111 
50|7753 12) 78621 7|796882/807304/817480 '827407|/337083 846503)/855665 864567 | 10 


511775496 786396|797057 807475 817648 82757 1|337242 846658/855816/864713| 9 
52/775679 786576) 797233 807647/817815 '827734|3837401/846813/855966 864860; 8 
53|775863| 786756) 797408 807818817982 |827897|337560 846967/856117/865006) 7 
o4 Ube. | Vetch 797584/807990/818150 |828060/837719)847122/856267|865151| 6 
55/776230 787114| 797759 808161818317 |828223| 337878 847277/356417 865297) 5 
56|776413 787294) 797935/808333/818484 828386] 338036 84743 1/856567 865443 4 
2 
1 
0 
M 


57|/776596! 787473} 798110/808504/818651 |828549| 338195 847585)856718 865589 
58/776780) 787652| 798285) 808675)/818818 |8287 12} 338354 847740/356868 865734 
59/776963|787832| 798460/808846)818985 |828875| 3385 12/847894/857017 865880 
60 777146 788011} 798636|809017|819152|829038) 338671 848048|357167|866025 


| 39° “39° | 38° | 37° | 36° | 35° | 34° | 38° | 32° | 31° | 30° | ™ 


Natural Co-sines. 


1|866171|874761 
2/866316 874902 
86646 1'875942 
866607875183 
866752.875324 


867331875886 
867476/87692% 
86762 1/876167 
867765/8763)7 


WOK! SCWOM HNP 


— 
on 


— 
© @ 


WD Wt] Ww 
Wwe! © 


| 26)869782/878261 


35/871071 


140/871784/880201 


44872354 
45/872496 88989 | 


49/873064/881441 


50/873 206/88 1578/839582 


866897 875465/883766 
867042/'875605/3883902 
867187 .875746/884038 


884145 
884581 
13|867910/876447/884717 
14|/368054/876587/884852 
115|863199 876727|884988 
16/8683 43/376867/885 123 
17/868487/877006/885258 
868632/877 146/88539 4 
868776/877286/835529 
868920)877425/885664 


869064/877565/885799 
869207\877704|885934 
869351/877844|886069 
24|869495|877983/886204 
25|869639|878 1221886338 
886473 
27/869926/878100/886608 
28/870969/878539 886742 
29/870212/878678|886876 
30/8703561878817/887011 


31/870499/878956/887145 
432/8706 42,879 95|/887279 
33/870785|879233)/887413 
34/870928)879372/887548 
8795 10/887681 
36/871214/879649/887815 
37/871357/879787|/887949 
138/871499|873925/888083 
139/87 1642'886063'8388217 
888350 
41871927 830339/883484 
142'872069 880477|/888617 
43872212 880615/888751 
880753/8888 34 
889017 
46|/872638 831028/889150 
47|872789/881166|889283 
48)/872922'881303'889416 


883084 
883221 
883357 
883493 
383624 


881174 
884309 


889549 


893894 
894024 
894154 
894284 
894415 
894545 
894675 
894805 
894934/902585 


895V64/902710 
895194 
895323 
895453 
895582 
895712 
395841 
895970 
896099 
896229 
896353 
896486 
896615 
896744 
896873 
3897001 
897130 


897387 
897515 


NATURAL SINES. 


891007 
891139 
591270 
891402 
491534 
891666 
891798 
891929 
892061 
892192 
892323 
892455 
392586 
392717 
3928 18 


892979 


393110 


89324 1/909951 
39337 1/90 1077|908508)9 15663 
393502/901203/908630)9 15779 
393633/901329/908751 
7991393 /63/901455|908872/9 16013 
901581/90899 4/9 16130 
901707\909 1 15)916246 


897258 


898794906358 913545, 
898922/906 4311913664 
899 349/906554/9 13782/929732 
899176/996676|913900 
899304/906799/914918 
906922)/914136 
899558/97044'9 14254 
899685/907166 914372 
8998 12/907289/9 14490 
899939/90741 1/9 14607 
900085/997533/914725 
9.)7655|9 14842 
9003191907777/9 14960 
900445/9N7899/915077 
915194 
900692/908143/915311 


899431 


900192 


900572/908021 


991958 
92084/909478/916595 
992299/909599)916712 
902335/90972019 16828 
992460/909841 916944 
999961 917060 
910082 917176 
902236910202 917292 
910323 917408 
903086/910443 917523 
9032101910563. 917639 
90333519 106841917755 
903460 910304 917870 
903585 910924917986 
903709911044/918101 
903834 911164/918216 
903958 911284/918331 
904083/911403/918446 
94207911523/918561 
911643/918676 
9445519 11762/918791 
918906 
919021 
904827 912120)919135 
994951/9 1223 11919950 
995075 912358919364 


990825/908265)915429 
915546 


908387 


902961 


915896 


920505 
920618 


920846 
920959 
921072 
921185 
921299 
921412 
921525 
921638 


921750 
921863 


921976 
922088 
922201 
922313 
922426 
922538 
922650 
922762 


922874 


m| 60° | 61° _62" | 63" | 64" 65° | 66° | 67° | 68° | 69° Im 
~0/866025/874620/882948 927184|933580/60 
927293'933685/59 
927402/933789/58 
927510/933393)57 
927619/933997\: 
927728)/934 101 
927836/934204/54 
927945/934308|53 
998053/934412 52 
928161 
928270 
928378/934722/49 
928486|/934826/48 
928594|934929/47 
928702/935034/46 
9288 10/935135/45 
928917 
929025/935341/43 
929133/935444/42 
929246 
9293 181935650/40 


999.455/935752139 


934515/5] 
434619|50 


935238\44 


935547/41 


922986)/929562)935855/38 
923998/929669|935957|37 
90 1833)}99923 6/9 16363)/92321 0/929776|936060/36 


909357/916479/923322/929884/936 162/35 


923434 /929990|935264/34 
923545/930097|936366|33 
923657/930204|936468)32 


923768/930311 


9238809304 18/936672/30 
92399 1/93052 493677429 


924102/939631 


936876)28 


924213|930737/936977/27 
92 4324|930843/937079)26 


924435)/930950/937181 


921546 /931056/937282/24 


924657 
924768 
924878 
924989 


931162, 937383)23 
931268|/937485/22 
931374/937586|21 
931480/937687)/20 


904331 


90457991188] 
904703)9 12001 


153/873631/881999 


54/873772)/882157/890213) 


5518739 14/8822 64 
56/874055 882401 
57/874196|882533 
58|874338 882674 
59/874479/8828 11 
60/874629/882948 


“29° | 28° 


8733147 881716889815 
52/873489'881853/889948 


890980 


890345 
890478 


890610/898411 
890742'898539 


890874 
891007 


er 


893028 


897643 
897771 


897900 
898156 
898283 


898666 
898794 


26° 


25° | 24° 


Natural Co-sines. 


995 193/912 177|919479 
905322 912596)919593 
905445/9 12715)9 19707 
905569)9 1283 1919821 
9056929 12953)9 19936 
905815)913972/920050, 
905939/913190/920164 
906062)913309|/929277 
996185/9 13427/920391 
906398/913545)920505 927184/933580/939693 


| 22° 


925099 
925210 
925321) 
92543) 
925541 
925651 
925761 
925871 
925980 
926090 
926200 
926310 
926419 
926529 


931586/937788 

931691 
931797/937990 
J31992)/93809 1 
932008|933191 
932113/938292 
9322 19|938393 
932324/938493 
932429/933593) 11 
932534/938694 
932639933794 
932744938894 
932849/933994 
932954/939094 
926638/933058 939194 
926747|933163)/939294 
926857/933267|939394 
926966/933372/939493 
927075|/933476/939593 


937889 


S— OW RAInwdwse 


ee SS 


“21° | 20° | m} 


2° 
“01939693 
11939792 
2/939891 
31939991 


945613)95114€| 956390 
945708/951236|956475 
'945802/951326|956560 
4/940090 945897 /951415|956644 
5|940189,945991|951505/956729 
6)940288 946085) 951594/956814 
7/940387/946180/951684) 956898 


73° | 


945519}951057| 956305 


74° 
961262 
961342 
961422 
961502 
961582 
861662 
961741 
961821 


75° 
965926 
966001 
966076 
966151 
966226 
966301 
966376 
966451 


76° 
970296 
970366 
970436 
970506 
970577 
970647 
970716 
970786 


974370) 
974435 
974501 
974566 
974631 
974696 
974761 
974826 


77° | 78° | 79° im 


Y (5148) ISL027 
978208 |I81683 
978268 |J81738 
978329 |I81793 
978389 |981849 
978449 |981904 
978509 |981959 
978569 |982014 


60 
59 
38 
97 
56 
95 
D4 
53 


81940486 
9/940585 
10940684 
111940782) 
12/940881 
940979 
941078 
941176 
941274 
941372 
941471 
19/941569 
20/941666 


21\941764 
22)941862 
23'941960 
24/942057 
25)942155 
26/942252 
27|942350 


946274/951773 
946368|951862 
946462|951951 
946555)952U40 
946649 |952129 
946743 !952218 
946837|952307 
946930/952396 
947024/952484 
947117|952573 
947210)952661 
947304/952750 
947397/952838 


947490|952926 
947583/953015 
947676|953103 
947768/953191 
947861/953279 
947954|953366 
948046/953454 
948139|953542 


28/942447 


961901 
961980 
962059 


96213y/966749 


966526 
966600 
966675 


956983 
957067 
957151 


957235 


9573 19|}962218)966823 
957404/962297/966898 
957487|962376|966972 
957571/962455 967046 
957655|962534/967120 
957739 |962613/967194 
957822 /962692/967268 
957906|962770|967342 
957990|962849 967415 


958073 |962928'967489 
958156|963006)/967562 
958239963084 967636 
958323 /963163/967709 
958406/963241/967782 
958489 |963319/967856 


971273 
971342 
971411 
971480 


971687 


971755 
971824 
971893 
971961 
972030 | 
972098 


974891 |978629 
970926 |974956 |978689 
970995 |975020 |978748 


971065 1975085 |9 1S8Us 
971134 '975149 |J78867 
971204 975214 |978927 
975278 |978986 
975342 1979045 
975406 |979105 
975471 |979164 
975535 (979223 
1975598 |979282 
975662 979341 


975726 |979399 
975790 979458 
975853 |9 79517 
975917979575 
975980 |979634 
976044 |J79692 


970856 


971549 
971618 


958572)963397|967929 


972166 
958654/963475/968002'972234 
958737 |963553|968075 972302 


976107 | 179750 
976170 |979809 
976233 |979367 


982069 
982123 
982178 
932233 
982287 
982342 
982396 
982450 
982505144 
982559143 
982613/49 
982667/41 
982721/40 


982774 < 
982828 
982882 
982935 
982980 
983042): 
983096): 
983149|< 
983202 


52 
5] 
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40/943609 
41/943705 
42/943801 
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46/944185 
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49 944472 
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949059|954414|959478 
959560 
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959724 
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959968 
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960131 

960212 
960294 
960875 
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961181 
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948324 


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949334)954674 
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949517|954847 
949608|954934 
949699/955020 
949790|955106 
949881/955192 
949972|955278 
950063|955364 
950154/955450 
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950516)955793 
950606|955879 
950696) 955964 
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950877)956134 
950967) 956220 


60/945519 


m} 19° | 18° | 17° | 16° 


951057|956305 


958820/963630/968148 972370 


958902|963708/968220 972438 
963786|968293 972506 
963863/968366 972573 976485 |930098 
963941/968438 
964018 
964095 
964173 
964250 
964327 
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964481 
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970014|974108 
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970155|974239 
9702251974305, 
9659261970296 974370 


-_— | | 


15° | 14° | 138° 


| 


Natural Co-sines. 


97796 


976296 |979925 


976359 |979983 983308/29 


976422) 980041 


976547 |980156 
976610 |980214 
9766721980271 
976735 |980329 
976797 |980386 
976850 |980443 
976921 |980500 


976984 980558 
977046 /980615 
977108|980672 
977169 |980728 
977231 |980785 
977293 |980842 
977354 |380899 
977416 |980955 
977477 |981012 
977539 |981068 
977600 |Y81124 
977661 |981181 
977722 |981237 
977783 |981293 
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931460 
981516 
981572 
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984298) 10 
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16,985605|988106 99 39 05|993103/994998996589/9.17878 998862/999542) 999918\44 
17}935654/988450 990944|993137 935027 9966 14|997597 998876/999551) 999922/43 
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22/985899]988669|991183}093306)095170.9967321997990)398944)999594 99993938 
23985947 938713 991177/993339 | 998008/998957/999602| 9999 49/37 
24/9599 6 938756991216 |993373 995227 99 '779|99'3027 998971/999610) 999945 36 
95/988045)988300 991254) )93406| 395256 996802 998015 993984999618, 999948 35 
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31/98633 1 989 Jo9991485/993605 995424 995940 998153 999061999165, 999964)29 
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ERRATA. 


BOOK I. In the diagram to Prop. II, a right line should be 
drawn fron Fito A. | 

In Prop. VI, the parallel lines kK, sl-ould be drawn parallel 
to the axis EF. 

In Scholium to Prop. VII, AB or CD, should be the axis 
of the parabola instead of AC or BD. 

On wes 31, eighth line from bottom, read (dz—z’) instead of 


BOOK II. Page 60, second line from bottom, IN should be 
changed to LN. 

Page 61, third line from the top, MH should be changed to 
GLMH. 

BOOK III. On page 101, eleventh line from the bottom, for 
fluction, read fluxion. 

On page 110. eight lines from the bottom, for rz, read ars, 
and for 2is read 3rs. 

Page 112, sixteenth line from bottom, for rs read rs. yee 
thirteenth line from bottom, supply 2 to 3rs, to make it read 3 ors. 

Page 113 Formula 2, for 15978 read }sz.; and in the next line 


for #rs read °rs. 


On pages 143 and 144, CQ should be changed to AQ CP. 
to AP, and fifth line from bottom, page 144, change C to A; also 
in the third line from top for »n/ar—z’ “Pts rA/ 2az—z’. 

BOOK Y. On 158 page Article 6th fourth line, for zz read 
zz; and on the fifth line, for z= read z~; also in article “7th on 
the ninth line, for J (dx) x, tead a/(dx)z. | 

Page 158, tenth line from top, for AD, read BC; alsoin article 
10th, ‘supply the inferior dash to z, in first, second, third and 
fourth lines, so that they may read z’, az’, az and az’; also on 
the sixth line, article 10th, for an/z, ead nite 

Page 152, on tenth line, for au, read au. 


y 


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. 


UNIVERSITY OF ILLINOIS-URBANA 


516SCH6H C001 
HIGHER GEOMETRY AND TRIGONOMETRY NEW YO 


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